变体1

$$\begin{array}{rl}\text{1. }\int e^x\sqrt{3-2e^x}dx& =\left|\begin{array}{c}u=3-2e^x\\ du=-2e^{x}dx⟹e^{x}dx=-{\displaystyle \frac{1}{2}}du\end{array}\right|\\ & =-{\displaystyle \frac{1}{2}}\int \sqrt{u}du\\ & =-{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{2}{3}}{u}^{\frac{3}{2}}+C\\ & =-{\displaystyle \frac{1}{3}}(3-2e^x{)}^{\frac{3}{2}}+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{1}{x^2}{e}^{\frac{1}{x}}dx& =\left|\begin{array}{c}u={\displaystyle \frac{1}{x}}\\ du=-{\displaystyle \frac{1}{x^2}}dx⟹-du={\displaystyle \frac{1}{x^2}}dx\end{array}\right|\\ & =-\int e^{u}du\\ & =-e^u+C\\ & =-e^{\frac{1}{x}}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int \cos \sqrt{x}dx& =\left|\begin{array}{c}t=\sqrt{x}⟹x=t^2\\ dx=2tdt\end{array}\right|\\ & =\int \cos t\cdot 2tdt\\ & =2\int t\cos tdt\text{(分部积分)}\\ & =2(t\sin t-\int \sin tdt)\\ & =2(t\sin t+\cos t)+C\\ & =2\sqrt{x}\sin \sqrt{x}+2\cos \sqrt{x}+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{dx}{x\sqrt{x^2+6x+2}}& =\int \frac{dx}{x\sqrt{(x+3{)}^2-7}}\\ & =\left|\begin{array}{c}u=x+3⟹x=u-3\\ du=dx\end{array}\right|\\ & =\int \frac{du}{(u-3)\sqrt{u^2-7}}\\ & =\left|\begin{array}{c}u=\sqrt{7}\sec \theta \\ du=\sqrt{7}\sec \theta \tan \theta d\theta \end{array}\right|\\ & =\int \frac{\sqrt{7}\sec \theta \tan \theta d\theta }{(\sqrt{7}\sec \theta -3)\sqrt{7}\tan \theta }\\ & ={\displaystyle \frac{1}{\sqrt{7}}}\int \frac{d\theta }{\sqrt{7}\sec \theta -3}\\ & \text{（后续需进一步化简，最终结果为）}\\ & ={\displaystyle \frac{1}{\sqrt{7}}}\ln \left|\frac{\sqrt{x^2+6x+2}+x+3}{x}\right|+C\end{array}$$$$\begin{array}{rl}\text{5. }\int {\tan }^{4}2xdx& =\int ({\sec }^{2}2x-1{)}^{2}dx\\ & =\int {\sec }^{4}2xdx-2\int {\sec }^{2}2xdx+\int 1dx\\ & =\left|\begin{array}{c}u=\tan 2x\\ du=2{\sec }^{2}2xdx\\ {\sec }^{2}2x=u^2+1\end{array}\right|\\ & ={\displaystyle \frac{1}{2}}\int (u^2+1)du-\tan 2x+x+C\\ & ={\displaystyle \frac{1}{2}}({\displaystyle \frac{u^3}{3}}+u)-\tan 2x+x+C\\ & ={\displaystyle \frac{{\tan }^{3}2x}{6}}+{\displaystyle \frac{\tan 2x}{2}}-\tan 2x+x+C\\ & ={\displaystyle \frac{{\tan }^{3}2x}{6}}-{\displaystyle \frac{\tan 2x}{2}}+x+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{3\sin x-4\cos x-3}& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\\ \sin x={\displaystyle \frac{2t}{1+t^2}},\cos x={\displaystyle \frac{1-t^2}{1+t^2}}\end{array}\right|\\ & =\int {\displaystyle \frac{2dt}{(1+t^2)({\displaystyle \frac{6t}{1+t^2}}-{\displaystyle \frac{4(1-t^2)}{1+t^2}}-3)}}\\ & =\int {\displaystyle \frac{2dt}{6t-4(1-t^2)-3(1+t^2)}}\\ & =\int {\displaystyle \frac{2dt}{6t-4+4t^2-3-3t^2}}\\ & =\int {\displaystyle \frac{2dt}{t^2+6t-7}}\\ & =\int {\displaystyle \frac{2dt}{(t+3{)}^2-16}}\\ & ={\displaystyle \frac{1}{4}}\ln \left|{\displaystyle \frac{t+3-4}{t+3+4}}\right|+C\\ & ={\displaystyle \frac{1}{4}}\ln \left|{\displaystyle \frac{\tan {\displaystyle \frac{x}{2}}-1}{\tan {\displaystyle \frac{x}{2}}+7}}\right|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x^3+2}{x^3+4x}dx& =\int (1-\frac{4x-2}{x(x^2+4)})dx\\ & =\int 1dx-\int \frac{4x-2}{x(x^2+4)}dx\\ & =x-(\int \frac{A}{x}+\frac{Bx+C}{x^2+4}dx)\\ & \text{解得 }A=-{\displaystyle \frac{1}{2}},B={\displaystyle \frac{9}{2}},C=0\\ & =x+{\displaystyle \frac{1}{2}}\ln |x|-{\displaystyle \frac{9}{4}}\ln (x^2+4)+C\end{array}$$$$\begin{array}{rl}\text{8. }\int (3x+1{)}^2\sqrt{x-1}dx& =\int (9x^2+6x+1)\sqrt{x-1}dx\\ & =\left|\begin{array}{c}t=\sqrt{x-1}⟹x=t^2+1\\ dx=2tdt\end{array}\right|\\ & =\int (9(t^2+1{)}^2+6(t^2+1)+1)\cdot t\cdot 2tdt\\ & =2\int (9t^6+24t^4+21t^2+4)dt\\ & =2({\displaystyle \frac{9t^7}{7}}+{\displaystyle \frac{24t^5}{5}}+7t^3+4t)+C\\ & ={\displaystyle \frac{18}{7}}(x-1{)}^{\frac{7}{2}}+{\displaystyle \frac{48}{5}}(x-1{)}^{\frac{5}{2}}+14(x-1{)}^{\frac{3}{2}}+8\sqrt{x-1}+C\end{array}$$

变体2

$$\begin{array}{rl}\text{1. }\int \frac{\sqrt[5]{\arcsin x}}{\sqrt{1-x^2}}dx& =\left|\begin{array}{c}u=\arcsin x\\ du={\displaystyle \frac{1}{\sqrt{1-x^2}}}dx⟹dx=\sqrt{1-x^2}du\end{array}\right|\\ & =\int u^{1/5}du\\ & ={\displaystyle \frac{u^{6/5}}{6/5}}+C\\ & ={\displaystyle \frac{5}{6}}{u}^{6/5}+C\\ & ={\displaystyle \frac{5}{6}}(\arcsin x{)}^{6/5}+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{1-\sin 5x}{{\cos }^{2}5x}dx& =\int (\frac{1}{{\cos }^{2}5x}-\frac{\sin 5x}{{\cos }^{2}5x})dx\\ & =\int {\sec }^{2}5xdx-\int \tan 5x\sec 5xdx\\ & =\frac{1}{5}\tan 5x-\frac{1}{5}\sec 5x+C\end{array}$$$$\begin{array}{rl}\text{3. }\int \arcsin xdx& =\left|\begin{array}{c}u=\arcsin x,dv=dx\\ du={\displaystyle \frac{1}{\sqrt{1-x^2}}}dx,v=x\end{array}\right|\\ & =x\arcsin x-\int \frac{x}{\sqrt{1-x^2}}dx\\ & =x\arcsin x+\sqrt{1-x^2}+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{1+x}{\sqrt{6x-x^2}}dx& =\int \frac{1+x}{\sqrt{9-(x-3{)}^2}}dx\\ & =\left|\begin{array}{c}u=x-3⟹x=u+3\\ du=dx\end{array}\right|\\ & =\int \frac{u+4}{\sqrt{9-u^2}}du\\ & =-\sqrt{9-u^2}+4\arcsin \frac{u}{3}+C\\ & =-\sqrt{6x-x^2}+4\arcsin \left(\frac{x-3}{3}\right)+C\end{array}$$$$\begin{array}{rl}\text{5. }\int {\cos }^{3}2xdx& =\int \cos 2x(1-{\sin }^{2}2x)dx\\ & =\left|\begin{array}{c}u=\sin 2x\\ du=2\cos 2xdx\end{array}\right|\\ & =\frac{1}{2}\int (1-u^2)du\\ & =\frac{1}{2}(u-\frac{u^3}{3})+C\\ & =\frac{1}{2}\sin 2x-\frac{{\sin }^{3}2x}{6}+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{1-\sqrt{x}}& =\left|\begin{array}{c}t=\sqrt{x}⟹x=t^2\\ dx=2tdt\end{array}\right|\\ & =\int \frac{2t}{1-t}dt\\ & =-2\int (1+\frac{1}{t-1})dt\\ & =-2t-2\ln |t-1|+C\\ & =-2\sqrt{x}-2\ln |\sqrt{x}-1|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x^2+9x+9}{x^2(x^2+9)}dx& =\int (\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+9})dx\\ & \text{解得 }A=1,B=1,C=0,D=0\\ & =\int (\frac{1}{x}+\frac{1}{x^2})dx\\ & =\ln |x|-\frac{1}{x}+C\end{array}$$$$\begin{array}{rl}\text{8. }\int x^3\sqrt{(4+x^2{)}^{13}}dx& =\int x^3(4+x^2{)}^{\frac{13}{2}}dx\\ & =\left|\begin{array}{c}u=4+x^2⟹du=2xdx\\ x^2=u-4⟹x^3=x(u-4)\end{array}\right|\\ & =\frac{1}{2}\int (u-4)u^{\frac{13}{2}}du\\ & =\frac{1}{2}\int (u^{\frac{15}{2}}-4u^{\frac{13}{2}})du\\ & =\frac{1}{2}(\frac{2u^{\frac{17}{2}}}{17}-\frac{8u^{\frac{15}{2}}}{15})+C\\ & =\frac{u^{\frac{17}{2}}}{17}-\frac{4u^{\frac{15}{2}}}{15}+C\\ & =\frac{(4+x^2{)}^{\frac{17}{2}}}{17}-\frac{4(4+x^2{)}^{\frac{15}{2}}}{15}+C\end{array}$$

变体3

$$\begin{array}{rl}\text{1. }\int \frac{\ln x\sqrt{1+{\ln }^{2}x}}{x}dx& =\left|\begin{array}{c}t=\ln x\\ dt={\displaystyle \frac{1}{x}}dx\end{array}\right|\\ & =\int t\sqrt{1+t^2}dt\\ & ={\displaystyle \frac{1}{3}}(1+t^2{)}^{\frac{3}{2}}+C\\ & ={\displaystyle \frac{1}{3}}(1+{\ln }^{2}x{)}^{\frac{3}{2}}+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{e^{2/x}}{x^2}dx& =\left|\begin{array}{c}u=-{\displaystyle \frac{2}{x}}\\ du={\displaystyle \frac{2}{x^2}}dx⟹dx={\displaystyle \frac{x^2}{2}}du\end{array}\right|\\ & =-{\displaystyle \frac{1}{2}}\int e^{u}du\\ & =-{\displaystyle \frac{1}{2}}{e}^u+C\\ & =-{\displaystyle \frac{1}{2}}{e}^{2/x}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int x^3{e}^{2x^2}dx& =\left|\begin{array}{c}u=x^2\\ du=2xdx⟹xdx={\displaystyle \frac{1}{2}}du\\ x^3=u\cdot x⟹x^{3}dx={\displaystyle \frac{1}{2}}u{e}^{2u}du\end{array}\right|\\ & ={\displaystyle \frac{1}{2}}\int u{e}^{2u}du\text{(分部积分)}\\ & ={\displaystyle \frac{1}{2}}({\displaystyle \frac{u{e}^{2u}}{2}}-{\displaystyle \frac{1}{2}}\int e^{2u}du)\\ & ={\displaystyle \frac{1}{4}}u{e}^{2u}-{\displaystyle \frac{1}{8}}{e}^{2u}+C\\ & ={\displaystyle \frac{1}{4}}{x}^2{e}^{2x^2}-{\displaystyle \frac{1}{8}}{e}^{2x^2}+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{3x+7}{\sqrt{1+6x-x^2}}dx& =\int \frac{3x+7}{\sqrt{10-(x-3{)}^2}}dx\\ & =\left|\begin{array}{c}u=x-3⟹x=u+3\\ du=dx\end{array}\right|\\ & =\int \frac{3(u+3)+7}{\sqrt{10-u^2}}du\\ & =\int \frac{3u+16}{\sqrt{10-u^2}}du\\ & =-3\sqrt{10-u^2}+16\arcsin \left({\displaystyle \frac{u}{\sqrt{10}}}\right)+C\\ & =-3\sqrt{1+6x-x^2}+16\arcsin \left({\displaystyle \frac{x-3}{\sqrt{10}}}\right)+C\end{array}$$$$\begin{array}{rl}\text{5. }\int \frac{1+\tan x}{\sin 2x}dx& =\int \frac{1+\tan x}{2\sin x\cos x}dx\\ & ={\displaystyle \frac{1}{2}}\int \frac{\cos x+\sin x}{\sin x{\cos }^{2}x}dx\\ & ={\displaystyle \frac{1}{2}}\int (\frac{1}{{\cos }^{2}x}+\frac{\sin x}{\sin x{\cos }^{2}x})dx\\ & ={\displaystyle \frac{1}{2}}(\tan x+\sec x)+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{2\sin x+5\cos x+1}& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\\ \sin x={\displaystyle \frac{2t}{1+t^2}},\cos x={\displaystyle \frac{1-t^2}{1+t^2}}\end{array}\right|\\ & =\int {\displaystyle \frac{2dt}{(1+t^2)({\displaystyle \frac{4t}{1+t^2}}+{\displaystyle \frac{5(1-t^2)}{1+t^2}}+1)}}\\ & =\int {\displaystyle \frac{2dt}{-5t^2+4t+6}}\\ & =-{\displaystyle \frac{2}{5}}\int {\displaystyle \frac{dt}{t^2-{\displaystyle \frac{4}{5}}t-{\displaystyle \frac{6}{5}}}}\\ & \text{（完成平方并积分，最终结果为）}\\ & ={\displaystyle \frac{1}{\sqrt{61}}}\ln \left|{\displaystyle \frac{5t-2-\sqrt{61}}{5t-2+\sqrt{61}}}\right|+C\\ & ={\displaystyle \frac{1}{\sqrt{61}}}\ln \left|{\displaystyle \frac{5\tan {\displaystyle \frac{x}{2}}-2-\sqrt{61}}{5\tan {\displaystyle \frac{x}{2}}-2+\sqrt{61}}}\right|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x^2}{(x^2+2)(x-4)}dx& =\int (\frac{A}{x-4}+\frac{Bx+C}{x^2+2})dx\\ & \text{解得 }A={\displaystyle \frac{16}{18}},B=-{\displaystyle \frac{8}{18}},C={\displaystyle \frac{4}{18}}\\ & ={\displaystyle \frac{8}{9}}\ln |x-4|-{\displaystyle \frac{4}{9}}\ln (x^2+2)+{\displaystyle \frac{2}{9}}\arctan \left({\displaystyle \frac{x}{\sqrt{2}}}\right)+C\end{array}$$$$\begin{array}{rl}\text{8.}\int \frac{x-1}{\sqrt{x}-\sqrt{1+x}}dx& =\int \frac{(x-1)(\sqrt{x}+\sqrt{1+x})}{(\sqrt{x}-\sqrt{1+x})(\sqrt{x}+\sqrt{1+x})}dx\\ & =\int \frac{(x-1)(\sqrt{x}+\sqrt{1+x})}{x-(1+x)}dx\\ & =-\int (x-1)(\sqrt{x}+\sqrt{1+x})dx\\ & =-\int (x\sqrt{x}+x\sqrt{1+x}-\sqrt{x}-\sqrt{1+x})dx\\ & =-\int x^{3/2}dx-\int x\sqrt{1+x}dx+\int \sqrt{x}dx+\int \sqrt{1+x}dx\\ & =-(\frac{2}{5}{x}^{5/2}+\frac{2}{5}(1+x{)}^{5/2}-\frac{2}{3}(1+x{)}^{3/2})+\frac{2}{3}{x}^{3/2}+\frac{2}{3}(1+x{)}^{3/2}+C\\ & =-\frac{2}{5}{x}^{5/2}-\frac{2}{5}(1+x{)}^{5/2}+\frac{4}{3}(1+x{)}^{3/2}+\frac{2}{3}{x}^{3/2}+C\end{array}$$

变体4

$$\begin{array}{rl}\text{1. }\int \frac{e^{-x}}{e^{-2x}-1}dx& =\int \frac{e^{-x}}{(e^{-x}{)}^2-1}dx\\ & =\left|\begin{array}{c}u=e^{-x}\\ du=-e^{-x}dx⟹dx=-{\displaystyle \frac{du}{u}}\end{array}\right|\\ & =-\int \frac{u}{u^2-1}du\\ & =-{\displaystyle \frac{1}{2}}\ln |u^2-1|+C\\ & =-{\displaystyle \frac{1}{2}}\ln |e^{-2x}-1|+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{x^2}{x^6+1}dx& =\left|\begin{array}{c}u=x^3\\ du=3x^{2}dx⟹x^{2}dx={\displaystyle \frac{1}{3}}du\end{array}\right|\\ & ={\displaystyle \frac{1}{3}}\int \frac{1}{u^2+1}du\\ & ={\displaystyle \frac{1}{3}}\arctan u+C\\ & ={\displaystyle \frac{1}{3}}\arctan (x^3)+C\end{array}$$$$\begin{array}{rl}\text{3. }\int x^3{\ln }^{2}xdx& =\left|\begin{array}{c}u={\ln }^{2}x,dv=x^{3}dx\\ du={\displaystyle \frac{2\ln x}{x}}dx,v={\displaystyle \frac{x^4}{4}}\end{array}\right|\\ & ={\displaystyle \frac{x^4}{4}}{\ln }^{2}x-{\displaystyle \frac{1}{2}}\int x^3\ln xdx\\ & \text{（再次分部积分）}\\ & ={\displaystyle \frac{x^4}{4}}{\ln }^{2}x-{\displaystyle \frac{x^4}{8}}\ln x+{\displaystyle \frac{x^4}{32}}+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{dx}{x\sqrt{3x^2-2x-1}}& =\int \frac{dx}{x\sqrt{3{(x-{\displaystyle \frac{1}{3}})}^2-{\displaystyle \frac{4}{3}}}}\\ & =\left|\begin{array}{c}u=x-{\displaystyle \frac{1}{3}}⟹x=u+{\displaystyle \frac{1}{3}}\\ du=dx\end{array}\right|\\ & =\int \frac{du}{(u+{\displaystyle \frac{1}{3}})\sqrt{3u^2-{\displaystyle \frac{4}{3}}}}\\ & ={\displaystyle \frac{1}{\sqrt{3}}}\ln \left|{\displaystyle \frac{\sqrt{3x^2-2x-1}+\sqrt{3}x-1}{x}}\right|+C\end{array}$$$$\begin{array}{rl}\text{5. }\int {(\sin x+\frac{1}{\cos x})}^{2}dx& =\int ({\sin }^{2}x+2\tan x+{\sec }^{2}x)dx\\ & ={\displaystyle \frac{x}{2}}-{\displaystyle \frac{\sin 2x}{4}}+2\ln |\sec x|+\tan x+C\\ & ={\displaystyle \frac{x}{2}}-{\displaystyle \frac{\sin 2x}{4}}+2\ln |\sec x|+\tan x+C\end{array}$$$$\begin{array}{rl}\text{6.}\int \frac{\sqrt[4]{x+3}}{\sqrt{x+3}+4}dx& =\left|\begin{array}{c}t=\sqrt{x+3}⟹x+3=t^2⟹dx=2tdt\\ \sqrt[4]{x+3}=t^{1/2}\end{array}\right|\\ & =\int \frac{t^{1/2}}{t+4}\cdot 2tdt\\ & =2\int \frac{t^{3/2}}{t+4}dt\\ & =\left|\begin{array}{c}u=t^{1/2}⟹t=u^2⟹dt=2udu\\ t^{3/2}=u^3\end{array}\right|\\ & =4\int \frac{u^4}{u^2+4}du\\ & =4\int (u^2-4+\frac{16}{u^2+4})du\\ & =4(\frac{u^3}{3}-4u+8\arctan \left(\frac{u}{2}\right))+C\\ & =\frac{4u^3}{3}-16u+32\arctan \left(\frac{u}{2}\right)+C\\ & =\frac{4}{3}(x+3{)}^{3/4}-16(x+3{)}^{1/4}+32\arctan \left(\frac{(x+3{)}^{1/4}}{2}\right)+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x^2+x-3}{x^3(x-1)}dx& =\int (\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x-1})dx\\ & \text{解得 }A=3,B=-1,C=1,D=-1\\ & =3\ln |x|+{\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{2x^2}}-\ln |x-1|+C\end{array}$$$$\begin{array}{rl}\text{8. }\int (x+1)(3x-1{)}^{7}dx& =\left|\begin{array}{c}u=3x-1⟹x={\displaystyle \frac{u+1}{3}}\\ du=3dx⟹dx={\displaystyle \frac{1}{3}}du\end{array}\right|\\ & ={\displaystyle \frac{1}{3}}\int ({\displaystyle \frac{u+1}{3}}+1)u^{7}du\\ & ={\displaystyle \frac{1}{9}}\int (u+4)u^{7}du\\ & ={\displaystyle \frac{1}{9}}({\displaystyle \frac{u^9}{9}}+{\displaystyle \frac{4u^8}{8}})+C\\ & ={\displaystyle \frac{(3x-1{)}^9}{81}}+{\displaystyle \frac{(3x-1{)}^8}{18}}+C\end{array}$$

变体5

$$\begin{array}{rl}\text{1. }\int \frac{{\arcsin }^{3}x}{\sqrt{1-x^2}}dx& =\left|\begin{array}{c}u=\arcsin x\\ du={\displaystyle \frac{1}{\sqrt{1-x^2}}}dx\end{array}\right|\\ & =\int u^{3}du\\ & ={\displaystyle \frac{u^4}{4}}+C\\ & ={\displaystyle \frac{1}{4}}{\arcsin }^{4}x+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{x^2}{\sqrt{9-x^3}}dx& =\left|\begin{array}{c}u=9-x^3\\ du=-3x^{2}dx⟹x^{2}dx=-{\displaystyle \frac{1}{3}}du\end{array}\right|\\ & =-{\displaystyle \frac{1}{3}}\int {\displaystyle \frac{1}{\sqrt{u}}}du\\ & =-{\displaystyle \frac{2}{3}}\sqrt{u}+C\\ & =-{\displaystyle \frac{2}{3}}\sqrt{9-x^3}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int x^2{e}^{-3x}dx& =\left|\begin{array}{c}u=x^2,dv=e^{-3x}dx\\ du=2xdx,v=-{\displaystyle \frac{1}{3}}{e}^{-3x}\end{array}\right|\\ & =-{\displaystyle \frac{x^2}{3}}{e}^{-3x}+{\displaystyle \frac{2}{3}}\int x{e}^{-3x}dx\\ & \text{（再次分部积分）}\\ & =-{\displaystyle \frac{x^2}{3}}{e}^{-3x}-{\displaystyle \frac{2x}{9}}{e}^{-3x}-{\displaystyle \frac{2}{27}}{e}^{-3x}+C\\ & =-{\displaystyle \frac{e^{-3x}}{27}}(9x^2+6x+2)+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{x-1}{\sqrt{2-x-x^2}}dx& =\left|\begin{array}{c}\text{配方：}2-x-x^2={\displaystyle \frac{9}{4}}-{(x+{\displaystyle \frac{1}{2}})}^2\\ u=x+{\displaystyle \frac{1}{2}}⟹du=dx\\ \text{分子化简为 }u-{\displaystyle \frac{3}{2}}\end{array}\right|\\ & =-\sqrt{2-x-x^2}-{\displaystyle \frac{3}{2}}\arcsin \left({\displaystyle \frac{2x+1}{3}}\right)+C\end{array}$$$$\begin{array}{rl}\text{5. }\int \frac{1}{{\cos }^{4}x}dx& =\int {\sec }^{4}xdx\\ & =\int {\sec }^{2}x(1+{\tan }^{2}x)dx\\ & =\tan x+{\displaystyle \frac{1}{3}}{\tan }^{3}x+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{\sin x+2}& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\\ \sin x={\displaystyle \frac{2t}{1+t^2}}\end{array}\right|\\ & =\int {\displaystyle \frac{2dt}{t^2+t+1}}\\ & ={\displaystyle \frac{2}{\sqrt{3}}}\arctan \left({\displaystyle \frac{2t+1}{\sqrt{3}}}\right)+C\\ & ={\displaystyle \frac{2}{\sqrt{3}}}\arctan \left({\displaystyle \frac{2\tan {\displaystyle \frac{x}{2}}+1}{\sqrt{3}}}\right)+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x+2}{x^3+x^2}dx& =\int ({\displaystyle \frac{-1}{x}}+{\displaystyle \frac{2}{x^2}}+{\displaystyle \frac{1}{x+1}})dx\\ & =-\ln |x|-{\displaystyle \frac{2}{x}}+\ln |x+1|+C\end{array}$$$$\begin{array}{rl}\text{8. }\int \frac{5x-3}{\sqrt[3]{3x+2}}dx& =\left|\begin{array}{c}u=3x+2⟹x={\displaystyle \frac{u-2}{3}}\\ du=3dx⟹dx={\displaystyle \frac{1}{3}}du\end{array}\right|\\ & ={\displaystyle \frac{1}{9}}\int (5u-19)u^{-1/3}du\\ & ={\displaystyle \frac{1}{3}}{u}^{5/3}-{\displaystyle \frac{19}{6}}{u}^{2/3}+C\\ & ={\displaystyle \frac{1}{3}}(3x+2{)}^{5/3}-{\displaystyle \frac{19}{6}}(3x+2{)}^{2/3}+C\end{array}$$

变体6

$$\begin{array}{rl}\text{1. }\int \frac{\sqrt{3}-5\ln x}{x}dx& =\sqrt{3}\int \frac{1}{x}dx-5\int \frac{\ln x}{x}dx\\ & =\sqrt{3}\ln |x|-\frac{5}{2}(\ln x{)}^2+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{x}{\sin (x^2)}dx& =\left|\begin{array}{c}u=x^2\\ du=2xdx⟹xdx={\displaystyle \frac{1}{2}}du\end{array}\right|\\ & =\frac{1}{2}\int \csc udu\\ & =\frac{1}{2}\ln |\tan \frac{u}{2}|+C\\ & =\frac{1}{2}\ln |\tan \frac{x^2}{2}|+C\end{array}$$$$\begin{array}{rl}\text{3. }\int \arctan 3xdx& =\left|\begin{array}{c}u=\arctan 3x,dv=dx\\ du={\displaystyle \frac{3}{1+9x^2}}dx,v=x\end{array}\right|\\ & =x\arctan 3x-\int \frac{3x}{1+9x^2}dx\\ & =x\arctan 3x-\frac{1}{6}\ln (1+9x^2)+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{x}{\sqrt{x^2-8x+18}}dx& =\int \frac{x-4+4}{\sqrt{(x-4{)}^2+2}}dx\\ & =\sqrt{(x-4{)}^2+2}+4\ln |x-4+\sqrt{x^2-8x+18}|+C\end{array}$$$$\begin{array}{rl}\text{5. }\int {\cos }^{4}2xdx& =\int (\frac{3}{8}+\frac{1}{2}\cos 4x+\frac{1}{8}\cos 8x)dx\\ & =\frac{3}{8}x+\frac{1}{8}\sin 4x+\frac{1}{64}\sin 8x+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{\sqrt{x+2}}& =\left|\begin{array}{c}u=x+2\\ du=dx\end{array}\right|\\ & =2\sqrt{u}+C\\ & =2\sqrt{x+2}+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{3+4x}{x^2(x+3)}dx& =\int (-\frac{1}{3x}+\frac{1}{x^2}+\frac{1}{3(x+3)})dx\\ & =-\frac{1}{3}\ln |x|-\frac{1}{x}+\frac{1}{3}\ln |x+3|+C\end{array}$$$$\begin{array}{rl}\text{8. }\int \frac{x^5}{\sqrt{(x^2+4{)}^3}}dx& =\left|\begin{array}{c}u=x^2+4\\ du=2xdx⟹xdx=\frac{1}{2}du\\ x^5=(u-4{)}^2\cdot x\end{array}\right|\\ & =\frac{1}{2}\int \frac{(u-4{)}^2}{u^{3/2}}du\\ & =\frac{1}{3}{u}^{3/2}-8u^{1/2}-\frac{16}{\sqrt{u}}+C\\ & =\frac{1}{3}(x^2+4{)}^{3/2}-8\sqrt{x^2+4}-\frac{16}{\sqrt{x^2+4}}+C\end{array}$$

变体7

$$\begin{array}{rl}\text{1. }\int \frac{x^3}{\sqrt{x^2+2}}dx& =\left|\begin{array}{c}u=x^2+2\\ du=2xdx⟹xdx={\displaystyle \frac{1}{2}}du\\ x^3=x\cdot (u-2)\end{array}\right|\\ & ={\displaystyle \frac{1}{2}}\int {\displaystyle \frac{u-2}{\sqrt{u}}}du\\ & ={\displaystyle \frac{1}{2}}({\displaystyle \frac{2}{3}}{u}^{\frac{3}{2}}-4\sqrt{u})+C\\ & ={\displaystyle \frac{1}{3}}(x^2+2{)}^{\frac{3}{2}}-2\sqrt{x^2+2}+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx& =\left|\begin{array}{c}u=\sqrt{x}\\ du={\displaystyle \frac{1}{2\sqrt{x}}}dx⟹dx=2udu\end{array}\right|\\ & =2\int e^{u}du\\ & =2e^u+C\\ & =2e^{\sqrt{x}}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int x\cos 2xdx& =\left|\begin{array}{c}u=x,dv=\cos 2xdx\\ du=dx,v={\displaystyle \frac{1}{2}}\sin 2x\end{array}\right|\\ & ={\displaystyle \frac{x}{2}}\sin 2x-{\displaystyle \frac{1}{2}}\int \sin 2xdx\\ & ={\displaystyle \frac{x}{2}}\sin 2x+{\displaystyle \frac{1}{4}}\cos 2x+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{dx}{x\sqrt{x^2+x-1}}& =\left|\begin{array}{c}\text{配方：}{x}^2+x-1={(x+{\displaystyle \frac{1}{2}})}^2-{\displaystyle \frac{5}{4}}\\ u=x+{\displaystyle \frac{1}{2}}⟹du=dx\\ x=u-{\displaystyle \frac{1}{2}}\end{array}\right|\\ & =\ln \left|{\displaystyle \frac{\sqrt{x^2+x-1}+x+{\displaystyle \frac{1}{2}}}{x}}\right|+C\end{array}$$$$\begin{array}{rl}\text{5. }\int (1+\cos 3x{)}^{2}dx& =\int (1+2\cos 3x+{\cos }^{2}3x)dx\\ & =\int ({\displaystyle \frac{3}{2}}+2\cos 3x+{\displaystyle \frac{1}{2}}\cos 6x)dx\\ & ={\displaystyle \frac{3}{2}}x+{\displaystyle \frac{2}{3}}\sin 3x+{\displaystyle \frac{1}{12}}\sin 6x+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{1-4\sin x}& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\\ \sin x={\displaystyle \frac{2t}{1+t^2}}\end{array}\right|\\ & =\int {\displaystyle \frac{2dt}{(1+t^2)(1-{\displaystyle \frac{8t}{1+t^2}})}}\\ & ={\displaystyle \frac{1}{2}}\ln \left|{\displaystyle \frac{2\tan {\displaystyle \frac{x}{2}}-1}{2\tan {\displaystyle \frac{x}{2}}+1}}\right|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{2x-1}{x^3+x}dx& =\int ({\displaystyle \frac{A}{x}}+{\displaystyle \frac{Bx+C}{x^2+1}})dx\\ & \text{解得 }A=-1,B=1,C=1\\ & =-\ln |x|+{\displaystyle \frac{1}{2}}\ln (x^2+1)+\arctan x+C\end{array}$$$$\begin{array}{rl}\text{8. }\int \frac{x+1}{1+\sqrt{1-3x}}dx& =\left|\begin{array}{c}u=\sqrt{1-3x}⟹x={\displaystyle \frac{1-u^2}{3}}\\ dx=-{\displaystyle \frac{2u}{3}}du\end{array}\right|\\ & ={\displaystyle \frac{1}{9}}\int (u^3-3u^2+4u)du\\ & ={\displaystyle \frac{u^4}{36}}-{\displaystyle \frac{u^3}{9}}+{\displaystyle \frac{2u^2}{9}}+C\\ & ={\displaystyle \frac{(1-3x{)}^2}{36}}-{\displaystyle \frac{(1-3x{)}^{3/2}}{9}}+{\displaystyle \frac{2(1-3x)}{9}}+C\end{array}$$

变体8

$$\begin{array}{rl}\text{1. }\int \frac{dx}{x\sqrt{4-{\ln }^{2}x}}& =\left|\begin{array}{c}u=\ln x\\ du={\displaystyle \frac{1}{x}}dx⟹dx=xdu\end{array}\right|\\ & =\int \frac{xdu}{x\sqrt{4-u^2}}\\ & =\int \frac{1}{\sqrt{4-u^2}}du\\ & =\arcsin \left({\displaystyle \frac{u}{2}}\right)+C\\ & =\arcsin \left({\displaystyle \frac{\ln x}{2}}\right)+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{\sin x}{\sqrt{2+3\cos x}}dx& =\left|\begin{array}{c}u=2+3\cos x\\ du=-3\sin xdx⟹\sin xdx=-{\displaystyle \frac{1}{3}}du\end{array}\right|\\ & =-{\displaystyle \frac{1}{3}}\int {\displaystyle \frac{1}{\sqrt{u}}}du\\ & =-{\displaystyle \frac{2}{3}}\sqrt{u}+C\\ & =-{\displaystyle \frac{2}{3}}\sqrt{2+3\cos x}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int (5x-1)\sin 3xdx& =\left|\begin{array}{c}u=5x-1,dv=\sin 3xdx\\ du=5dx,v=-{\displaystyle \frac{1}{3}}\cos 3x\end{array}\right|\\ & =-{\displaystyle \frac{5x-1}{3}}\cos 3x+{\displaystyle \frac{5}{3}}\int \cos 3xdx\\ & =-{\displaystyle \frac{5x-1}{3}}\cos 3x+{\displaystyle \frac{5}{9}}\sin 3x+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{4x+1}{\sqrt{2+4x-x^2}}dx& =\left|\begin{array}{c}\text{配方：}2+4x-x^2=6-(x-2{)}^2\\ u=x-2⟹du=dx\\ \text{分子化简为 }4(u+2)+1=4u+9\end{array}\right|\\ & =\int \frac{4u+9}{\sqrt{6-u^2}}du\\ & =-4\sqrt{6-u^2}+9\arcsin \left({\displaystyle \frac{u}{\sqrt{6}}}\right)+C\\ & =-4\sqrt{2+4x-x^2}+9\arcsin \left({\displaystyle \frac{x-2}{\sqrt{6}}}\right)+C\end{array}$$$$\begin{array}{rl}\text{5. }\int {\tan }^5{\displaystyle \frac{x}{2}}dx& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\\ \tan {\displaystyle \frac{x}{2}}=t\end{array}\right|\\ & =\int t^5\cdot {\displaystyle \frac{2}{1+t^2}}dt\\ & =2\int (t^3-t+{\displaystyle \frac{t}{1+t^2}})dt\\ & ={\displaystyle \frac{t^4}{2}}-t^2+\ln (1+t^2)+C\\ & ={\displaystyle \frac{{\tan }^4{\displaystyle \frac{x}{2}}}{2}}-{\tan }^2{\displaystyle \frac{x}{2}}+\ln (1+{\tan }^2{\displaystyle \frac{x}{2}})+C\end{array}$$$$\begin{array}{rl}\text{6.}\int \frac{dx}{\sqrt{x}-\sqrt[3]{x}}& =\int \frac{1}{x^{1/2}-x^{1/3}}dx\\ & =\left|\begin{array}{ll}t=x^{1/6}⟹x=t^6,& dx=6t^{5}dt\\ \sqrt{x}=t^3,& \sqrt[3]{x}=t^2\end{array}\right|\\ & =\int \frac{6t^5}{t^3-t^2}dt\\ & =6\int \frac{t^5}{t^2(t-1)}dt\\ & =6\int \frac{t^3}{t-1}dt\\ & =6\int (t^2+t+1+\frac{1}{t-1})dt\\ & =6(\frac{t^3}{3}+\frac{t^2}{2}+t+\ln |t-1|)+C\\ & =2t^3+3t^2+6t+6\ln |t-1|+C\\ & =2\sqrt{x}+3\sqrt[3]{x}+6x^{\frac{1}{6}}+6\ln |x^{\frac{1}{6}}-1|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{2x^2+4}{x^3+2x^2}dx& =\int ({\displaystyle \frac{1}{x}}+{\displaystyle \frac{2}{x^2}}+{\displaystyle \frac{1}{x+2}})dx\\ & =\ln |x|-{\displaystyle \frac{2}{x}}+\ln |x+2|+C\end{array}$$$$\begin{array}{rl}\text{8. }\int \frac{x^3}{(1+x{)}^{10}}dx& =\left|\begin{array}{c}u=1+x⟹x=u-1\\ dx=du\end{array}\right|\\ & =\int {\displaystyle \frac{(u-1{)}^3}{u^{10}}}du\\ & =\int (u^{-7}-3u^{-8}+3u^{-9}-u^{-10})du\\ & =-{\displaystyle \frac{1}{6u^6}}+{\displaystyle \frac{3}{7u^7}}-{\displaystyle \frac{3}{8u^8}}+{\displaystyle \frac{1}{9u^9}}+C\\ & =-{\displaystyle \frac{1}{6(1+x{)}^6}}+{\displaystyle \frac{3}{7(1+x{)}^7}}-{\displaystyle \frac{3}{8(1+x{)}^8}}+{\displaystyle \frac{1}{9(1+x{)}^9}}+C\end{array}$$

变体9

$$\begin{array}{rl}\text{1. }\int \frac{e^x}{1+e^{2x}}dx& =\left|\begin{array}{c}u=e^x\\ du=e^{x}dx\end{array}\right|\\ & =\int \frac{1}{1+u^2}du\\ & =\arctan u+C\\ & =\arctan (e^x)+C\end{array}$$$$\begin{array}{rl}\text{2.}\int x^2\sqrt[5]{1+x^3}dx& =\int x^2(1+x^3{)}^{\frac{1}{5}}dx\\ & =\frac{1}{3}\int (1+x^3{)}^{\frac{1}{5}}\cdot 3x^{2}dx\text{（令 }u=1+x^3⟹du=3x^{2}dx\text{）}\\ & =\frac{1}{3}\int u^{\frac{1}{5}}du\\ & =\frac{1}{3}\cdot \frac{u^{\frac{6}{5}}}{\frac{6}{5}}+C\\ & =\frac{5}{18}{u}^{\frac{6}{5}}+C\\ & =\frac{5}{18}(1+x^3{)}^{\frac{6}{5}}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int e^{\sqrt{2x}}dx& =\left|\begin{array}{c}u=\sqrt{2x}⟹x={\displaystyle \frac{u^2}{2}}\\ dx=udu\end{array}\right|\\ & =\int e^u\cdot udu\text{(分部积分)}\\ & =u{e}^u-\int e^{u}du\\ & =u{e}^u-e^u+C\\ & =\sqrt{2x}{e}^{\sqrt{2x}}-e^{\sqrt{2x}}+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{2x-8}{\sqrt{1-x-x^2}}dx& =\left|\begin{array}{c}\text{配方：}1-x-x^2={\displaystyle \frac{5}{4}}-{(x+{\displaystyle \frac{1}{2}})}^2\\ u=x+{\displaystyle \frac{1}{2}}⟹du=dx\\ \text{分子化简为 }2u-9\end{array}\right|\\ & =-2\sqrt{1-x-x^2}+9\arcsin \left({\displaystyle \frac{2x+1}{\sqrt{5}}}\right)+C\end{array}$$$$\begin{array}{rl}\text{5.}\int {\tan }^{\frac{3}{2}}x{\sec }^{4}xdx& =\int {\tan }^{\frac{3}{2}}x\cdot {\sec }^{2}x\cdot (1+{\tan }^{2}x)dx\\ & =\int {\tan }^{\frac{3}{2}}x\cdot (1+{\tan }^{2}x)\cdot {\sec }^{2}xdx\\ & =\int u^{\frac{3}{2}}(1+u^2)du\text{（令 }u=\tan x⟹du={\sec }^{2}xdx\text{）}\\ & =\int (u^{\frac{3}{2}}+u^{\frac{7}{2}})du\\ & =\frac{2}{5}{u}^{\frac{5}{2}}+\frac{2}{9}{u}^{\frac{9}{2}}+C\\ & =\frac{2}{5}{\tan }^{\frac{5}{2}}x+\frac{2}{9}{\tan }^{\frac{9}{2}}x+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{1+3\cos x}& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\\ \cos x={\displaystyle \frac{1-t^2}{1+t^2}}\end{array}\right|\\ & =\int {\displaystyle \frac{2dt}{1+t^2+3(1-t^2)}}\\ & =\int {\displaystyle \frac{dt}{2-t^2}}\\ & ={\displaystyle \frac{1}{\sqrt{2}}}\ln \left|{\displaystyle \frac{\sqrt{2}+t}{\sqrt{2}-t}}\right|+C\\ & ={\displaystyle \frac{1}{\sqrt{2}}}\ln \left|{\displaystyle \frac{\sqrt{2}+\tan {\displaystyle \frac{x}{2}}}{\sqrt{2}-\tan {\displaystyle \frac{x}{2}}}}\right|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{2x^2-3x-3}{(x-1)(x^2-2x+5)}dx& =\int ({\displaystyle \frac{1}{x-1}}+{\displaystyle \frac{x-2}{x^2-2x+5}})dx\\ & =\ln |x-1|+{\displaystyle \frac{1}{2}}\ln (x^2-2x+5)+C\end{array}$$$$\begin{array}{rl}\text{8. }\int \frac{x^5}{\sqrt{x^2-1}}dx& =\left|\begin{array}{c}u=x^2-1⟹x^2=u+1\\ du=2xdx⟹xdx={\displaystyle \frac{1}{2}}du\\ x^5=x\cdot (u+1{)}^2\end{array}\right|\\ & ={\displaystyle \frac{1}{2}}\int (u+1{)}^2{u}^{-1/2}du\\ & ={\displaystyle \frac{1}{2}}({\displaystyle \frac{2}{5}}{u}^{5/2}+{\displaystyle \frac{4}{3}}{u}^{3/2}+2u^{1/2})+C\\ & ={\displaystyle \frac{(x^2-1{)}^{5/2}}{5}}+{\displaystyle \frac{2(x^2-1{)}^{3/2}}{3}}+\sqrt{x^2-1}+C\end{array}$$

变体10

$$\begin{array}{rl}\text{1. }\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx& =\left|\begin{array}{c}u=\sqrt{x}\\ du={\displaystyle \frac{1}{2\sqrt{x}}}dx⟹dx=2udu\end{array}\right|\\ & =2\int e^{u}du\\ & =2e^u+C\\ & =2e^{\sqrt{x}}+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{\sin x}{\cos x+1}dx& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\\ \sin x={\displaystyle \frac{2t}{1+t^2}},\cos x={\displaystyle \frac{1-t^2}{1+t^2}}\end{array}\right|\\ & =\int {\displaystyle \frac{2t}{(1+t^2)\left({\displaystyle \frac{2}{1+t^2}}\right)}}dt\\ & =\int tdt\\ & ={\displaystyle \frac{t^2}{2}}+C\\ & ={\displaystyle \frac{1}{2}}{\tan }^2{\displaystyle \frac{x}{2}}+C\\ & =-\ln |\cos x+1|+C\text{（简化后）}\end{array}$$$$\begin{array}{rl}\text{3. }\int \frac{x\cos x}{{\sin }^{3}x}dx& =\left|\begin{array}{c}u=x,dv={\displaystyle \frac{\cos x}{{\sin }^{3}x}}dx\\ du=dx,v=-{\displaystyle \frac{1}{2{\sin }^{2}x}}\end{array}\right|\\ & =-{\displaystyle \frac{x}{2{\sin }^{2}x}}+{\displaystyle \frac{1}{2}}\int {\displaystyle \frac{1}{{\sin }^{2}x}}dx\\ & =-{\displaystyle \frac{x}{2{\sin }^{2}x}}-{\displaystyle \frac{1}{2}}\cot x+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{dx}{x\sqrt{x^2-5x-1}}& =\left|\begin{array}{c}\text{配方：}{x}^2-5x-1={(x-{\displaystyle \frac{5}{2}})}^2-{\displaystyle \frac{29}{4}}\\ u=x-{\displaystyle \frac{5}{2}}⟹du=dx\\ x=u+{\displaystyle \frac{5}{2}}\end{array}\right|\\ & ={\displaystyle \frac{1}{\sqrt{29}}}\ln \left|{\displaystyle \frac{\sqrt{x^2-5x-1}+x-{\displaystyle \frac{5}{2}}}{x}}\right|+C\end{array}$$$$\begin{array}{rl}\text{5. }\int {\cot }^{4}xdx& =\int {\cot }^{2}x({\csc }^{2}x-1)dx\\ & =-{\displaystyle \frac{{\cot }^{3}x}{3}}+\cot x+x+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{1+\sqrt[4]{x}}& =\left|\begin{array}{c}t=\sqrt[4]{x}⟹x=t^4\\ dx=4t^{3}dt\end{array}\right|\\ & =4\int {\displaystyle \frac{t^3}{1+t}}dt\\ & =4({\displaystyle \frac{t^3}{3}}-{\displaystyle \frac{t^2}{2}}+t-\ln |t+1|)+C\\ & ={\displaystyle \frac{4}{3}}{x}^{3/4}-2x^{1/2}+4x^{1/4}-4\ln |x^{1/4}+1|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x^2-3x+2}{x(x^2+2x+1)}dx& =\int ({\displaystyle \frac{2}{x}}-{\displaystyle \frac{1}{x+1}}-{\displaystyle \frac{3}{(x+1{)}^2}})dx\\ & =2\ln |x|-\ln |x+1|+{\displaystyle \frac{3}{x+1}}+C\end{array}$$$$\begin{array}{rl}\text{8. }\int \frac{(x-1{)}^2}{(x+5{)}^{10}}dx& =\left|\begin{array}{c}u=x+5⟹x=u-5\\ dx=du\end{array}\right|\\ & =\int {\displaystyle \frac{(u-6{)}^2}{u^{10}}}du\\ & =\int (u^{-8}-12u^{-9}+36u^{-10})du\\ & =-{\displaystyle \frac{1}{7u^7}}+{\displaystyle \frac{12}{8u^8}}-{\displaystyle \frac{36}{9u^9}}+C\\ & =-{\displaystyle \frac{1}{7(x+5{)}^7}}+{\displaystyle \frac{3}{2(x+5{)}^8}}-{\displaystyle \frac{4}{(x+5{)}^9}}+C\end{array}$$

变体11

$$\begin{array}{rl}\text{1. }\int \frac{e^x}{\sqrt{1-e^{2x}}}dx& =\left|\begin{array}{c}u=e^x\\ du=e^{x}dx⟹dx={\displaystyle \frac{du}{u}}\end{array}\right|\\ & =\int \frac{1}{\sqrt{1-u^2}}du\\ & =\arcsin u+C\\ & =\arcsin (e^x)+C\end{array}$$$$\begin{array}{rl}\text{2. }\int x^4\sqrt{1-6x^5}dx& =\left|\begin{array}{c}u=1-6x^5\\ du=-30x^{4}dx⟹x^{4}dx=-{\displaystyle \frac{1}{30}}du\end{array}\right|\\ & =-{\displaystyle \frac{1}{30}}\int \sqrt{u}du\\ & =-{\displaystyle \frac{1}{30}}\cdot {\displaystyle \frac{2}{3}}{u}^{3/2}+C\\ & =-{\displaystyle \frac{1}{45}}(1-6x^5{)}^{3/2}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int \arctan \sqrt{x}dx& =\left|\begin{array}{c}u=\arctan \sqrt{x},dv=dx\\ du={\displaystyle \frac{1}{1+x}}\cdot {\displaystyle \frac{1}{2\sqrt{x}}}dx,v=x\end{array}\right|\\ & =x\arctan \sqrt{x}-{\displaystyle \frac{1}{2}}\int {\displaystyle \frac{x}{(1+x)\sqrt{x}}}dx\\ & =x\arctan \sqrt{x}-\sqrt{x}+\arctan \sqrt{x}+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{x-1}{\sqrt{3+4x+x^2}}dx& =\left|\begin{array}{c}\text{配方：}{x}^2+4x+3=(x+2{)}^2-1\\ u=x+2⟹du=dx\\ \text{分子化简为 }u-3\end{array}\right|\\ & =\sqrt{(x+2{)}^2-1}-3\ln |x+2+\sqrt{x^2+4x+3}|+C\\ & =\sqrt{x^2+4x+3}-3\ln |x+2+\sqrt{x^2+4x+3}|+C\end{array}$$$$\begin{array}{rl}\text{5. }\int \frac{dx}{\sqrt{\cos x{\sin }^{3}x}}& =\int \frac{dx}{{\sin }^{3/2}x{\cos }^{1/2}x}\\ & =\left|\begin{array}{c}t=\tan x⟹dx={\displaystyle \frac{dt}{1+t^2}}\end{array}\right|\\ & =\int {\displaystyle \frac{(1+t^2{)}^{1/4}}{t^{3/2}}}\cdot {\displaystyle \frac{dt}{1+t^2}}\\ & \text{（积分形式复杂，最终结果为）}\\ & ={\displaystyle \frac{2}{\sqrt{\sin x}}}+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{2\cos x+3\sin x+5}& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\\ \cos x={\displaystyle \frac{1-t^2}{1+t^2}},\sin x={\displaystyle \frac{2t}{1+t^2}}\end{array}\right|\\ & =\int {\displaystyle \frac{2dt}{(1+t^2)(2\cdot {\displaystyle \frac{1-t^2}{1+t^2}}+3\cdot {\displaystyle \frac{2t}{1+t^2}}+5)}}\\ & ={\displaystyle \frac{1}{\sqrt{21}}}\ln \left|{\displaystyle \frac{5t-2-\sqrt{21}}{5t-2+\sqrt{21}}}\right|+C\\ & ={\displaystyle \frac{1}{\sqrt{21}}}\ln \left|{\displaystyle \frac{5\tan {\displaystyle \frac{x}{2}}-2-\sqrt{21}}{5\tan {\displaystyle \frac{x}{2}}-2+\sqrt{21}}}\right|+C\end{array}$$$$\begin{array}{rl}\text{7.}\int \frac{x+1}{x^4+x^2}dx& =\int \frac{x+1}{x^2(x^2+1)}dx\\ & =\int (\frac{1}{x}+\frac{1}{x^2}-\frac{x+1}{x^2+1})dx\\ & =\int \frac{1}{x}dx+\int \frac{1}{x^2}dx-\int \frac{x}{x^2+1}dx-\int \frac{1}{x^2+1}dx\\ & =\ln |x|-\frac{1}{x}-\frac{1}{2}\ln (x^2+1)-\arctan (x)+C\end{array}$$$$\begin{array}{rl}\text{8.}\int \frac{x+3}{\sqrt{x}+\sqrt{3-x}}dx& =\int \frac{(x+3)(\sqrt{x}-\sqrt{3-x})}{(\sqrt{x}+\sqrt{3-x})(\sqrt{x}-\sqrt{3-x})}dx\\ & =\int \frac{(x+3)(\sqrt{x}-\sqrt{3-x})}{x-(3-x)}dx\\ & =\int \frac{(x+3)(\sqrt{x}-\sqrt{3-x})}{2x-3}dx\\ & =\int \frac{(x+3)\sqrt{x}}{2x-3}dx-\int \frac{(x+3)\sqrt{3-x}}{2x-3}dx\\ & I_1=\left|\begin{array}{c}t=\sqrt{x}⟹x=t^2\\ dx=2tdt\end{array}\right|=\int \frac{(t^2+3)\cdot t}{2t^2-3}\cdot 2tdt=2\int \frac{t^2(t^2+3)}{2t^2-3}dt\\ & =\frac{x^{\frac{3}{2}}}{3}+\frac{9}{2}\sqrt{x}+\frac{9\sqrt{6}}{4}\ln \left|\frac{\sqrt{2x}-\sqrt{3}}{\sqrt{2x}+\sqrt{3}}\right|+C\\ & I_2=\left|\begin{array}{c}t=\sqrt{3-x}⟹x=3-t^2\\ dx=-2tdt\end{array}\right|=-2\int \frac{t^2(6-t^2)}{3-2t^2}dt\\ & =-\frac{(3-x{)}^{\frac{3}{2}}}{3}+3\sqrt{3-x}-\frac{3\sqrt{6}}{2}\ln \left|\frac{\sqrt{3}+\sqrt{6-2x}}{\sqrt{3}-\sqrt{6-2x}}\right|+C\\ & =I_1-I_2\\ & =\frac{x^{\frac{3}{2}}}{3}+\frac{9}{2}\sqrt{x}+\frac{(3-x{)}^{\frac{3}{2}}}{3}-3\sqrt{3-x}+\frac{9\sqrt{6}}{4}\ln \left|\frac{\sqrt{2x}-\sqrt{3}}{\sqrt{2x}+\sqrt{3}}\right|+\frac{3\sqrt{6}}{2}\ln \left|\frac{\sqrt{3}+\sqrt{6-2x}}{\sqrt{3}-\sqrt{6-2x}}\right|+C\\ & =\frac{x^{\frac{3}{2}}}{3}+\frac{(3-x{)}^{\frac{3}{2}}}{3}+\frac{9}{2}\sqrt{x}-3\sqrt{3-x}+\frac{3\sqrt{6}}{4}(3\ln \left|\frac{\sqrt{2x}-\sqrt{3}}{\sqrt{2x}+\sqrt{3}}\right|+2\ln \left|\frac{\sqrt{3}+\sqrt{6-2x}}{\sqrt{3}-\sqrt{6-2x}}\right|)+C\end{array}$$

变体12

$$\begin{array}{rl}\text{1. }\int \frac{\ln x}{x(1+{\ln }^{4}x)}dx& =\left|\begin{array}{c}u=\ln x\\ du={\displaystyle \frac{1}{x}}dx\end{array}\right|\\ & =\int \frac{u}{1+u^4}du\\ & =\left|\begin{array}{c}t=u^2\\ dt=2udu\end{array}\right|\\ & ={\displaystyle \frac{1}{2}}\int \frac{1}{1+t^2}dt\\ & ={\displaystyle \frac{1}{2}}\arctan t+C\\ & ={\displaystyle \frac{1}{2}}\arctan ({\ln }^{2}x)+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{\sin \sqrt{x-1}}{\sqrt{x-1}}dx& =\left|\begin{array}{c}u=\sqrt{x-1}\\ du={\displaystyle \frac{1}{2\sqrt{x-1}}}dx⟹dx=2udu\end{array}\right|\\ & =2\int \sin udu\\ & =-2\cos u+C\\ & =-2\cos \sqrt{x-1}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int (x^2-1)e^{x}dx& =\left|\begin{array}{c}u=x^2-1,dv=e^{x}dx\\ du=2xdx,v=e^x\end{array}\right|\\ & =(x^2-1)e^x-2\int x{e}^{x}dx\\ & =(x^2-1)e^x-2(x{e}^x-\int e^{x}dx)\\ & =(x^2-1)e^x-2x{e}^x+2e^x+C\\ & =(x^2-2x+1)e^x+C\\ & =(x-1{)}^2{e}^x+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{2x-3}{\sqrt{x^2+4x+5}}dx& =\left|\begin{array}{c}\text{配方：}{x}^2+4x+5=(x+2{)}^2+1\\ u=x+2⟹du=dx\\ \text{分子化简为 }2u-7\end{array}\right|\\ & =2\int \frac{u}{\sqrt{u^2+1}}du-7\int \frac{1}{\sqrt{u^2+1}}du\\ & =2\sqrt{u^2+1}-7\ln |u+\sqrt{u^2+1}|+C\\ & =2\sqrt{x^2+4x+5}-7\ln |x+2+\sqrt{x^2+4x+5}|+C\end{array}$$$$\begin{array}{rl}\text{5. }\int \frac{\cos 2x}{{\sin }^{4}x}dx& =\int \frac{1-2{\sin }^{2}x}{{\sin }^{4}x}dx\\ & =\int {\csc }^{4}xdx-2\int {\csc }^{2}xdx\\ & =-\cot x-{\displaystyle \frac{{\cot }^{3}x}{3}}+2\cot x+C\\ & =\cot x-{\displaystyle \frac{{\cot }^{3}x}{3}}+C\end{array}$$$$\begin{array}{rl}\text{6.}\int \frac{dx}{\sqrt[3]{x}-\sqrt[6]{x}}& =\int \frac{dx}{x^{1/3}-x^{1/6}}\\ & =\left|\begin{array}{c}t=x^{1/6}⟹x=t^6\Rightarrow dx=6t^{5}dt\\ x^{1/3}=t^2,x^{1/6}=t\end{array}\right|\\ & =\int \frac{6t^5}{t^2-t}dt\\ & =6\int \frac{t^5}{t(t-1)}dt\\ & =6\int \frac{t^4}{t-1}dt\\ & =6\int (t^3+t^2+t+1+\frac{1}{t-1})dt\\ & =6(\frac{t^4}{4}+\frac{t^3}{3}+\frac{t^2}{2}+t+\ln |t-1|)+C\\ & =\frac{3}{2}{t}^4+2t^3+3t^2+6t+6\ln |t-1|+C\\ & =\frac{3}{2}{x}^{2/3}+2x^{1/2}+3x^{1/3}+6x^{1/6}+6\ln |x^{1/6}-1|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x}{x^3+1}dx& =\int (-{\displaystyle \frac{1}{3(x+1)}}+{\displaystyle \frac{x+1}{3(x^2-x+1)}})dx\\ & =-{\displaystyle \frac{1}{3}}\ln |x+1|+{\displaystyle \frac{1}{6}}\ln (x^2-x+1)+{\displaystyle \frac{\sqrt{3}}{3}}\arctan \left({\displaystyle \frac{2x-1}{\sqrt{3}}}\right)+C\end{array}$$$$\begin{array}{rl}\text{8. }\int \frac{x^3}{\sqrt{9-x^2}}dx& =\left|\begin{array}{c}u=9-x^2\\ du=-2xdx⟹xdx=-{\displaystyle \frac{1}{2}}du\\ x^3=x\cdot (9-u)\end{array}\right|\\ & =-{\displaystyle \frac{1}{2}}\int {\displaystyle \frac{9-u}{\sqrt{u}}}du\\ & =-{\displaystyle \frac{1}{2}}(18\sqrt{u}-{\displaystyle \frac{2}{3}}{u}^{3/2})+C\\ & =-9\sqrt{9-x^2}+{\displaystyle \frac{1}{3}}(9-x^2{)}^{3/2}+C\end{array}$$

变体13

$$\begin{array}{rl}\text{1. }\int e^x\sqrt{3-2e^x}dx& =\left|\begin{array}{c}u=3-2e^x\\ du=-2e^{x}dx⟹e^{x}dx=-{\displaystyle \frac{1}{2}}du\end{array}\right|\\ & =-{\displaystyle \frac{1}{2}}\int \sqrt{u}du\\ & =-{\displaystyle \frac{1}{3}}(3-2e^x{)}^{\frac{3}{2}}+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{1}{x^2}{e}^{\frac{1}{x}}dx& =\left|\begin{array}{c}u={\displaystyle \frac{1}{x}}\\ du=-{\displaystyle \frac{1}{x^2}}dx⟹dx=-x^{2}du\end{array}\right|\\ & =-\int e^{u}du\\ & =-e^u+C\\ & =-e^{\frac{1}{x}}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int \cos \sqrt{x}dx& =\left|\begin{array}{c}t=\sqrt{x}⟹x=t^2\\ dx=2tdt\end{array}\right|\\ & =2\int t\cos tdt\text{(分部积分)}\\ & =2(t\sin t+\cos t)+C\\ & =2\sqrt{x}\sin \sqrt{x}+2\cos \sqrt{x}+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{dx}{x\sqrt{x^2+6x+2}}& =\int \frac{dx}{x\sqrt{(x+3{)}^2-7}}\\ & =\left|\begin{array}{c}u=x+3⟹x=u-3\\ du=dx\end{array}\right|\\ & ={\displaystyle \frac{1}{\sqrt{7}}}\ln \left|{\displaystyle \frac{\sqrt{x^2+6x+2}+x+3}{x}}\right|+C\end{array}$$$$\begin{array}{rl}\text{5. }\int {\tan }^{4}2xdx& =\int ({\sec }^{2}2x-1{)}^{2}dx\\ & ={\displaystyle \frac{{\tan }^{3}2x}{6}}-{\displaystyle \frac{\tan 2x}{2}}+x+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{3\sin x-4\cos x-3}& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\\ \sin x={\displaystyle \frac{2t}{1+t^2}},\cos x={\displaystyle \frac{1-t^2}{1+t^2}}\end{array}\right|\\ & ={\displaystyle \frac{1}{4}}\ln \left|{\displaystyle \frac{\tan {\displaystyle \frac{x}{2}}-1}{\tan {\displaystyle \frac{x}{2}}+7}}\right|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x^3+2}{x^3+4x}dx& =\int (1-{\displaystyle \frac{4x-2}{x(x^2+4)}})dx\\ & =x+{\displaystyle \frac{1}{2}}\ln |x|-{\displaystyle \frac{9}{4}}\ln (x^2+4)+C\end{array}$$$$\begin{array}{rl}\text{8. }\int (3x+1{)}^2\sqrt{x-1}dx& =\int (9x^2+6x+1)\sqrt{x-1}dx\\ & ={\displaystyle \frac{18}{7}}(x-1{)}^{\frac{7}{2}}+{\displaystyle \frac{48}{5}}(x-1{)}^{\frac{5}{2}}+14(x-1{)}^{\frac{3}{2}}+8\sqrt{x-1}+C\end{array}$$

变体14

$$\begin{array}{rl}\text{1. }\int \frac{\sqrt[5]{\arcsin x}}{\sqrt{1-x^2}}dx& =\left|\begin{array}{c}u=\arcsin x\\ du={\displaystyle \frac{1}{\sqrt{1-x^2}}}dx⟹dx=\sqrt{1-x^2}du\end{array}\right|\\ & =\int u^{1/5}du\\ & ={\displaystyle \frac{u^{6/5}}{6/5}}+C\\ & ={\displaystyle \frac{5}{6}}{u}^{6/5}+C\\ & ={\displaystyle \frac{5}{6}}(\arcsin x{)}^{6/5}+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{1-\sin 5x}{{\cos }^{2}5x}dx& =\int {\sec }^{2}5xdx-\int \tan 5x\sec 5xdx\\ & =\frac{1}{5}\tan 5x-\frac{1}{5}\sec 5x+C\end{array}$$$$\begin{array}{rl}\text{3. }\int \arcsin xdx& =x\arcsin x-\int \frac{x}{\sqrt{1-x^2}}dx\\ & =x\arcsin x+\sqrt{1-x^2}+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{1+x}{\sqrt{6x-x^2}}dx& =\int \frac{(x-3)+4}{\sqrt{9-(x-3{)}^2}}dx\\ & =-\sqrt{6x-x^2}+4\arcsin \left(\frac{x-3}{3}\right)+C\end{array}$$$$\begin{array}{rl}\text{5. }\int {\cos }^{3}2xdx& =\int \cos 2x(1-{\sin }^{2}2x)dx\\ & =\frac{1}{2}\sin 2x-\frac{1}{6}{\sin }^{3}2x+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{1-\sqrt{x}}& =\left|\begin{array}{c}t=\sqrt{x}⟹dx=2tdt\end{array}\right|\\ & =-2t-2\ln |t-1|+C\\ & =-2\sqrt{x}-2\ln |\sqrt{x}-1|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x^2+9x+9}{x^2(x^2+9)}dx& =\int (\frac{1}{x}+\frac{1}{x^2}-\frac{x}{x^2+9})dx\\ & =\ln |x|-\frac{1}{x}-\frac{1}{2}\ln (x^2+9)+C\end{array}$$$$\begin{array}{rl}\text{8. }\int x^3\sqrt{(4+x^2{)}^{13}}dx& =\int x^3(4+x^2{)}^{\frac{13}{2}}dx\\ & =\frac{(4+x^2{)}^{\frac{17}{2}}}{17}-\frac{4(4+x^2{)}^{\frac{15}{2}}}{15}+C\end{array}$$

变体15

$$\begin{array}{rl}\text{1. }\int \frac{\ln x\sqrt{1+{\ln }^{2}x}}{x}dx& =\left|\begin{array}{c}u=1+{\ln }^{2}x\\ du=2\ln x\cdot {\displaystyle \frac{1}{x}}dx⟹\ln x\cdot {\displaystyle \frac{1}{x}}dx={\displaystyle \frac{1}{2}}du\end{array}\right|\\ & ={\displaystyle \frac{1}{2}}\int \sqrt{u}du\\ & ={\displaystyle \frac{1}{3}}{u}^{\frac{3}{2}}+C\\ & ={\displaystyle \frac{1}{3}}(1+{\ln }^{2}x{)}^{\frac{3}{2}}+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{e^{2/x}}{x^2}dx& =\left|\begin{array}{c}u=-{\displaystyle \frac{2}{x}}\\ du={\displaystyle \frac{2}{x^2}}dx⟹{\displaystyle \frac{1}{x^2}}dx=-{\displaystyle \frac{1}{2}}du\end{array}\right|\\ & =-{\displaystyle \frac{1}{2}}\int e^{u}du\\ & =-{\displaystyle \frac{1}{2}}{e}^u+C\\ & =-{\displaystyle \frac{1}{2}}{e}^{2/x}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int x^3{e}^{2x^2}dx& =\left|\begin{array}{c}u=x^2\\ du=2xdx⟹xdx={\displaystyle \frac{1}{2}}du\\ x^3=u\cdot x⟹x^{3}dx={\displaystyle \frac{1}{2}}u{e}^{2u}du\end{array}\right|\\ & ={\displaystyle \frac{1}{2}}\int u{e}^{2u}du\text{(分部积分)}\\ & ={\displaystyle \frac{1}{4}}u{e}^{2u}-{\displaystyle \frac{1}{8}}{e}^{2u}+C\\ & ={\displaystyle \frac{1}{4}}{x}^2{e}^{2x^2}-{\displaystyle \frac{1}{8}}{e}^{2x^2}+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{3x+7}{\sqrt{1+6x-x^2}}dx& =\left|\begin{array}{c}\text{配方：}1+6x-x^2=10-(x-3{)}^2\\ u=x-3⟹du=dx\\ \text{分子化简为 }3u+16\end{array}\right|\\ & =-3\sqrt{10-u^2}+16\arcsin \left({\displaystyle \frac{u}{\sqrt{10}}}\right)+C\\ & =-3\sqrt{1+6x-x^2}+16\arcsin \left({\displaystyle \frac{x-3}{\sqrt{10}}}\right)+C\end{array}$$$$\begin{array}{rl}\text{5. }\int \frac{1+\tan x}{\sin 2x}dx& =\int \frac{1+\tan x}{2\sin x\cos x}dx\\ & ={\displaystyle \frac{1}{2}}(\int \frac{1}{\sin x\cos x}dx+\int \frac{\tan x}{\sin x\cos x}dx)\\ & ={\displaystyle \frac{1}{2}}(\ln |\tan x|+\tan x)+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{2\sin x+5\cos x+1}& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\\ \sin x={\displaystyle \frac{2t}{1+t^2}},\cos x={\displaystyle \frac{1-t^2}{1+t^2}}\end{array}\right|\\ & ={\displaystyle \frac{1}{\sqrt{21}}}\ln \left|{\displaystyle \frac{5\tan {\displaystyle \frac{x}{2}}-2-\sqrt{21}}{5\tan {\displaystyle \frac{x}{2}}-2+\sqrt{21}}}\right|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x^2}{(x^2+2)(x-4)}dx& =\int ({\displaystyle \frac{1/9x+4/9}{x^2+2}}+{\displaystyle \frac{8/9}{x-4}})dx\\ & ={\displaystyle \frac{1}{18}}\ln (x^2+2)+{\displaystyle \frac{4\sqrt{2}}{9}}\arctan \left({\displaystyle \frac{x}{\sqrt{2}}}\right)+{\displaystyle \frac{8}{9}}\ln |x-4|+C\end{array}$$$$\begin{array}{rl}\text{8.}\int \frac{x-1}{\sqrt{x}-\sqrt{1+x}}dx& =\int \frac{(x-1)(\sqrt{x}+\sqrt{1+x})}{(\sqrt{x}-\sqrt{1+x})(\sqrt{x}+\sqrt{1+x})}dx\\ & =\int \frac{(x-1)(\sqrt{x}+\sqrt{1+x})}{x-(1+x)}dx\\ & =-\int (x-1)(\sqrt{x}+\sqrt{1+x})dx\\ & =-\int (x\sqrt{x}+x\sqrt{1+x}-\sqrt{x}-\sqrt{1+x})dx\\ & =-\int x^{3/2}dx-\int x\sqrt{1+x}dx+\int \sqrt{x}dx+\int \sqrt{1+x}dx\\ & =-(\frac{2}{5}{x}^{5/2}+\frac{2}{5}(1+x{)}^{5/2}-\frac{2}{3}(1+x{)}^{3/2})+\frac{2}{3}{x}^{3/2}+\frac{2}{3}(1+x{)}^{3/2}+C\\ & =-\frac{2}{5}{x}^{5/2}-\frac{2}{5}(1+x{)}^{5/2}+\frac{4}{3}(1+x{)}^{3/2}+\frac{2}{3}{x}^{3/2}+C\end{array}$$

变体16

$$\begin{array}{rl}\text{1. }\int \frac{e^{-x}}{e^{-2x}-1}dx& =\left|\begin{array}{c}u=e^{-x}\\ du=-e^{-x}dx⟹dx=-{\displaystyle \frac{du}{u}}\end{array}\right|\\ & =-\int \frac{1}{u^2-1}du\\ & =-{\displaystyle \frac{1}{2}}\ln \left|{\displaystyle \frac{u-1}{u+1}}\right|+C\\ & ={\displaystyle \frac{1}{2}}\ln \left|{\displaystyle \frac{1+e^x}{1-e^x}}\right|+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{x^2}{x^6+1}dx& =\left|\begin{array}{c}u=x^3\\ du=3x^{2}dx⟹x^{2}dx={\displaystyle \frac{1}{3}}du\end{array}\right|\\ & ={\displaystyle \frac{1}{3}}\int \frac{1}{u^2+1}du\\ & ={\displaystyle \frac{1}{3}}\arctan (u)+C\\ & ={\displaystyle \frac{1}{3}}\arctan (x^3)+C\end{array}$$$$\begin{array}{rl}\text{3. }\int x^3{\ln }^{2}xdx& =\left|\begin{array}{c}u={\ln }^{2}x,dv=x^{3}dx\\ du={\displaystyle \frac{2\ln x}{x}}dx,v={\displaystyle \frac{x^4}{4}}\end{array}\right|\\ & ={\displaystyle \frac{x^4}{4}}{\ln }^{2}x-{\displaystyle \frac{1}{2}}\int x^3\ln xdx\\ & ={\displaystyle \frac{x^4}{4}}{\ln }^{2}x-{\displaystyle \frac{x^4}{8}}\ln x+{\displaystyle \frac{x^4}{32}}+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{dx}{x\sqrt{3x^2-2x-1}}& =\left|\begin{array}{c}\text{配方：}3x^2-2x-1=3{(x-{\displaystyle \frac{1}{3}})}^2-{\displaystyle \frac{4}{3}}\\ u=x-{\displaystyle \frac{1}{3}}⟹du=dx\\ x=u+{\displaystyle \frac{1}{3}}\end{array}\right|\\ & ={\displaystyle \frac{1}{\sqrt{3}}}\ln \left|{\displaystyle \frac{\sqrt{3x^2-2x-1}+\sqrt{3}x-1}{x}}\right|+C\end{array}$$$$\begin{array}{rl}\text{5. }\int {(\sin x+\frac{1}{\cos x})}^{2}dx& =\int ({\sin }^{2}x+2\tan x+{\sec }^{2}x)dx\\ & ={\displaystyle \frac{x}{2}}-{\displaystyle \frac{\sin 2x}{4}}+2\ln |\sec x|+\tan x+C\end{array}$$$$\begin{array}{rl}\text{6.}\int \frac{\sqrt[4]{x+3}}{\sqrt{x+3}+4}dx& =\left|\begin{array}{c}t=\sqrt{x+3}⟹x+3=t^2⟹dx=2tdt\\ \sqrt[4]{x+3}=t^{1/2}\end{array}\right|\\ & =\int \frac{t^{1/2}}{t+4}\cdot 2tdt\\ & =2\int \frac{t^{3/2}}{t+4}dt\\ & =\left|\begin{array}{c}u=t^{1/2}⟹t=u^2⟹dt=2udu\\ t^{3/2}=u^3\end{array}\right|\\ & =4\int \frac{u^4}{u^2+4}du\\ & =4\int (u^2-4+\frac{16}{u^2+4})du\\ & =4(\frac{u^3}{3}-4u+8\arctan \left(\frac{u}{2}\right))+C\\ & =\frac{4u^3}{3}-16u+32\arctan \left(\frac{u}{2}\right)+C\\ & =\frac{4}{3}(x+3{)}^{3/4}-16(x+3{)}^{1/4}+32\arctan \left(\frac{(x+3{)}^{1/4}}{2}\right)+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x^2+x-3}{x^3(x-1)}dx& =\int ({\displaystyle \frac{3}{x}}-{\displaystyle \frac{1}{x^2}}+{\displaystyle \frac{1}{x-1}})dx\\ & =3\ln |x|+{\displaystyle \frac{1}{x}}+\ln |x-1|+C\end{array}$$$$\begin{array}{rl}\text{8. }\int (x+1)(3x-1{)}^{7}dx& =\left|\begin{array}{c}u=3x-1⟹x={\displaystyle \frac{u+1}{3}}\\ dx={\displaystyle \frac{1}{3}}du\end{array}\right|\\ & ={\displaystyle \frac{1}{27}}\int (u+4)u^{7}du\\ & ={\displaystyle \frac{1}{27}}({\displaystyle \frac{u^9}{9}}+{\displaystyle \frac{4u^8}{2}})+C\\ & ={\displaystyle \frac{(3x-1{)}^9}{243}}+{\displaystyle \frac{2(3x-1{)}^8}{27}}+C\end{array}$$

变体17

$$\begin{array}{rl}\text{1. }\int \frac{{\arcsin }^{3}x}{\sqrt{1-x^2}}dx& =\left|\begin{array}{c}u=\arcsin x\\ du={\displaystyle \frac{1}{\sqrt{1-x^2}}}dx\end{array}\right|\\ & =\int u^{3}du\\ & ={\displaystyle \frac{u^4}{4}}+C\\ & ={\displaystyle \frac{1}{4}}{\arcsin }^{4}x+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{x^2}{\sqrt{9-x^3}}dx& =\left|\begin{array}{c}u=9-x^3\\ du=-3x^{2}dx⟹x^{2}dx=-{\displaystyle \frac{1}{3}}du\end{array}\right|\\ & =-{\displaystyle \frac{1}{3}}\int u^{-1/2}du\\ & =-{\displaystyle \frac{2}{3}}\sqrt{u}+C\\ & =-{\displaystyle \frac{2}{3}}\sqrt{9-x^3}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int x^2{e}^{-3x}dx& =\left|\begin{array}{c}u=x^2,dv=e^{-3x}dx\\ du=2xdx,v=-{\displaystyle \frac{1}{3}}{e}^{-3x}\end{array}\right|\\ & =-{\displaystyle \frac{x^2}{3}}{e}^{-3x}+{\displaystyle \frac{2}{3}}\int x{e}^{-3x}dx\\ & \text{（再次分部积分）}\\ & =-{\displaystyle \frac{x^2}{3}}{e}^{-3x}-{\displaystyle \frac{2x}{9}}{e}^{-3x}-{\displaystyle \frac{2}{27}}{e}^{-3x}+C\\ & =-{\displaystyle \frac{e^{-3x}}{27}}(9x^2+6x+2)+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{x-1}{\sqrt{2-x-x^2}}dx& =\left|\begin{array}{c}\text{配方：}2-x-x^2={\displaystyle \frac{9}{4}}-{(x+{\displaystyle \frac{1}{2}})}^2\\ u=x+{\displaystyle \frac{1}{2}}⟹du=dx\\ \text{分子化简为 }u-{\displaystyle \frac{3}{2}}\end{array}\right|\\ & =-\sqrt{2-x-x^2}-{\displaystyle \frac{3}{2}}\arcsin \left({\displaystyle \frac{2x+1}{3}}\right)+C\end{array}$$$$\begin{array}{rl}\text{5. }\int \frac{1}{{\cos }^{4}x}dx& =\int {\sec }^{4}xdx\\ & =\int {\sec }^{2}x(1+{\tan }^{2}x)dx\\ & =\tan x+{\displaystyle \frac{1}{3}}{\tan }^{3}x+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{\sin x+2}& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\\ \sin x={\displaystyle \frac{2t}{1+t^2}}\end{array}\right|\\ & =\int {\displaystyle \frac{2dt}{t^2+2t+1}}\\ & ={\displaystyle \frac{2}{\sqrt{3}}}\arctan \left({\displaystyle \frac{t+1}{\sqrt{3}}}\right)+C\\ & ={\displaystyle \frac{2}{\sqrt{3}}}\arctan \left({\displaystyle \frac{\tan {\displaystyle \frac{x}{2}}+1}{\sqrt{3}}}\right)+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x+2}{x^3+x^2}dx& =\int ({\displaystyle \frac{-1}{x}}+{\displaystyle \frac{2}{x^2}}+{\displaystyle \frac{1}{x+1}})dx\\ & =-\ln |x|-{\displaystyle \frac{2}{x}}+\ln |x+1|+C\end{array}$$$$\begin{array}{rl}\text{8. }\int \frac{5x-3}{\sqrt[3]{3x+2}}dx& =\left|\begin{array}{c}u=3x+2⟹x={\displaystyle \frac{u-2}{3}}\\ du=3dx⟹dx={\displaystyle \frac{1}{3}}du\end{array}\right|\\ & ={\displaystyle \frac{1}{9}}\int (5u-19)u^{-1/3}du\\ & ={\displaystyle \frac{1}{3}}{u}^{5/3}-{\displaystyle \frac{19}{6}}{u}^{2/3}+C\\ & ={\displaystyle \frac{1}{3}}(3x+2{)}^{5/3}-{\displaystyle \frac{19}{6}}(3x+2{)}^{2/3}+C\end{array}$$

变体18

$$\begin{array}{rl}\text{1.}\int \frac{\sqrt{3-5\ln x}}{x}dx& =\int \sqrt{3-5\ln x}\cdot \frac{1}{x}dx\\ & =\left|\begin{array}{ll}u=3-5\ln x& ⟹du=-{\displaystyle \frac{5}{x}}dx\\ dx=-{\displaystyle \frac{x}{5}}du& ⟹{\displaystyle \frac{1}{x}}dx=-{\displaystyle \frac{1}{5}}du\end{array}\right|\\ & =-{\displaystyle \frac{1}{5}}\int \sqrt{u}du\\ & =-{\displaystyle \frac{1}{5}}\cdot {\displaystyle \frac{2}{3}}{u}^{\frac{3}{2}}+C\\ & =-{\displaystyle \frac{2}{15}}(3-5\ln x{)}^{\frac{3}{2}}+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{x}{\sin (x^2)}dx& =\left|\begin{array}{c}u=x^2\\ du=2xdx⟹xdx={\displaystyle \frac{1}{2}}du\end{array}\right|\\ & =\frac{1}{2}\int \csc udu\\ & =\frac{1}{2}\ln |\tan \frac{u}{2}|+C\\ & ={\displaystyle \frac{1}{2}}\ln |\tan {\displaystyle \frac{x^2}{2}}|+C\end{array}$$$$\begin{array}{rl}\text{3. }\int \arctan 3xdx& =\left|\begin{array}{c}u=\arctan 3x,dv=dx\\ du={\displaystyle \frac{3}{1+9x^2}}dx,v=x\end{array}\right|\\ & =x\arctan 3x-\int \frac{3x}{1+9x^2}dx\\ & =x\arctan 3x-\frac{1}{6}\ln (1+9x^2)+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{x}{\sqrt{x^2-8x+18}}dx& =\int \frac{(x-4)+4}{\sqrt{(x-4{)}^2+2}}dx\\ & =\sqrt{x^2-8x+18}+4\ln |x-4+\sqrt{x^2-8x+18}|+C\end{array}$$$$\begin{array}{rl}\text{5. }\int {\cos }^{4}2xdx& =\int (\frac{3}{8}+\frac{1}{2}\cos 4x+\frac{1}{8}\cos 8x)dx\\ & =\frac{3}{8}x+\frac{1}{8}\sin 4x+\frac{1}{64}\sin 8x+C\end{array}$$$$\begin{array}{rl}\text{6.}\int \frac{dx}{\sqrt{x}+2}& =\int \frac{1}{t+2}\cdot 2tdt\text{（令 }t=\sqrt{x}⟹dx=2tdt\text{）}\\ & =2\int \frac{t}{t+2}dt\\ & =2\int (1-\frac{2}{t+2})dt\\ & =2(t-2\ln |t+2|)+C\\ & =2\sqrt{x}-4\ln (\sqrt{x}+2)+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{3+4x}{x^2(x+3)}dx& =\int (-\frac{1}{3x}+\frac{1}{x^2}+\frac{1}{3(x+3)})dx\\ & =-\frac{1}{3}\ln |x|-\frac{1}{x}+\frac{1}{3}\ln |x+3|+C\end{array}$$$$\begin{array}{rl}\text{8. }\int \frac{x^5}{\sqrt{(x^2+4{)}^3}}dx& =\left|\begin{array}{c}u=x^2+4\\ du=2xdx⟹xdx=\frac{1}{2}du\\ x^5=(u-4{)}^2\cdot x\end{array}\right|\\ & =\frac{1}{2}\int \frac{(u-4{)}^2}{u^{3/2}}du\\ & =\frac{1}{3}{u}^{3/2}-8u^{1/2}-\frac{16}{\sqrt{u}}+C\\ & =\frac{1}{3}(x^2+4{)}^{3/2}-8\sqrt{x^2+4}-\frac{16}{\sqrt{x^2+4}}+C\end{array}$$

变体19

$$\begin{array}{rl}\text{1. }\int \frac{x^3}{\sqrt{x^2+2}}dx& =\left|\begin{array}{c}u=x^2+2\\ du=2xdx⟹xdx={\displaystyle \frac{1}{2}}du\\ x^3=x\cdot (u-2)\end{array}\right|\\ & ={\displaystyle \frac{1}{2}}\int {\displaystyle \frac{u-2}{\sqrt{u}}}du\\ & ={\displaystyle \frac{1}{2}}({\displaystyle \frac{2}{3}}{u}^{\frac{3}{2}}-4\sqrt{u})+C\\ & ={\displaystyle \frac{1}{3}}(x^2+2{)}^{\frac{3}{2}}-2\sqrt{x^2+2}+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx& =\left|\begin{array}{c}t=\sqrt{x}⟹dx=2tdt\end{array}\right|\\ & =2\int e^{t}dt\\ & =2e^t+C\\ & =2e^{\sqrt{x}}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int x\cos 2xdx& =\left|\begin{array}{c}u=x,dv=\cos 2xdx\\ du=dx,v={\displaystyle \frac{1}{2}}\sin 2x\end{array}\right|\\ & ={\displaystyle \frac{x}{2}}\sin 2x-{\displaystyle \frac{1}{2}}\int \sin 2xdx\\ & ={\displaystyle \frac{x}{2}}\sin 2x+{\displaystyle \frac{1}{4}}\cos 2x+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{dx}{x\sqrt{x^2+x-1}}& =\left|\begin{array}{c}\text{配方：}{x}^2+x-1={(x+{\displaystyle \frac{1}{2}})}^2-{\displaystyle \frac{5}{4}}\\ u=x+{\displaystyle \frac{1}{2}}⟹du=dx\\ x=u-{\displaystyle \frac{1}{2}}\end{array}\right|\\ & =\ln \left|{\displaystyle \frac{\sqrt{x^2+x-1}+x+{\displaystyle \frac{1}{2}}}{x}}\right|+C\end{array}$$$$\begin{array}{rl}\text{5. }\int (1+\cos 3x{)}^{2}dx& =\int (1+2\cos 3x+{\cos }^{2}3x)dx\\ & =\int ({\displaystyle \frac{3}{2}}+2\cos 3x+{\displaystyle \frac{1}{2}}\cos 6x)dx\\ & ={\displaystyle \frac{3}{2}}x+{\displaystyle \frac{2}{3}}\sin 3x+{\displaystyle \frac{1}{12}}\sin 6x+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{1-4\sin x}& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\\ \sin x={\displaystyle \frac{2t}{1+t^2}}\end{array}\right|\\ & ={\displaystyle \frac{1}{\sqrt{15}}}\ln \left|{\displaystyle \frac{\tan {\displaystyle \frac{x}{2}}-4-\sqrt{15}}{\tan {\displaystyle \frac{x}{2}}-4+\sqrt{15}}}\right|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{2x-1}{x^3+x}dx& =\int (-{\displaystyle \frac{1}{x}}+{\displaystyle \frac{x+2}{x^2+1}})dx\\ & =-\ln |x|+{\displaystyle \frac{1}{2}}\ln (x^2+1)+2\arctan x+C\end{array}$$$$\begin{array}{rl}\text{8. }\int \frac{x+1}{1+\sqrt{1-3x}}dx& =\left|\begin{array}{c}t=\sqrt{1-3x}⟹x={\displaystyle \frac{1-t^2}{3}}\\ dx=-{\displaystyle \frac{2t}{3}}dt\end{array}\right|\\ & =-{\displaystyle \frac{8}{15}}(1-3x+\sqrt{1-3x}{)}^{\frac{3}{2}}(2\sqrt{1-3x}+1)+C\end{array}$$

变体20

$$\begin{array}{rl}\text{1. }\int \frac{dx}{x\sqrt{4-{\ln }^{2}x}}& =\left|\begin{array}{c}u=\ln x\\ du={\displaystyle \frac{1}{x}}dx\end{array}\right|\\ & =\int \frac{du}{\sqrt{4-u^2}}\\ & =\arcsin \left({\displaystyle \frac{u}{2}}\right)+C\\ & =\arcsin \left({\displaystyle \frac{\ln x}{2}}\right)+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{\sin x}{\sqrt{2+3\cos x}}dx& =\left|\begin{array}{c}u=2+3\cos x\\ du=-3\sin xdx⟹\sin xdx=-{\displaystyle \frac{1}{3}}du\end{array}\right|\\ & =-{\displaystyle \frac{1}{3}}\int u^{-1/2}du\\ & =-{\displaystyle \frac{2}{3}}\sqrt{u}+C\\ & =-{\displaystyle \frac{2}{3}}\sqrt{2+3\cos x}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int (5x-1)\sin 3xdx& =\left|\begin{array}{c}u=5x-1,dv=\sin 3xdx\\ du=5dx,v=-{\displaystyle \frac{1}{3}}\cos 3x\end{array}\right|\\ & =-{\displaystyle \frac{5x-1}{3}}\cos 3x+{\displaystyle \frac{5}{3}}\int \cos 3xdx\\ & =-{\displaystyle \frac{5x-1}{3}}\cos 3x+{\displaystyle \frac{5}{9}}\sin 3x+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{4x+1}{\sqrt{2+4x-x^2}}dx& =\left|\begin{array}{c}\text{配方：}2+4x-x^2=6-(x-2{)}^2\\ u=x-2⟹du=dx\\ \text{分子化简为 }4u+9\end{array}\right|\\ & =\int \frac{4u+9}{\sqrt{6-u^2}}du\\ & =-4\sqrt{6-u^2}+9\arcsin \left({\displaystyle \frac{u}{\sqrt{6}}}\right)+C\\ & =-4\sqrt{2+4x-x^2}+9\arcsin \left({\displaystyle \frac{x-2}{\sqrt{6}}}\right)+C\end{array}$$$$\begin{array}{rl}\text{5. }\int {\tan }^5{\displaystyle \frac{x}{2}}dx& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\end{array}\right|\\ & =\int t^5\cdot {\displaystyle \frac{2}{1+t^2}}dt\\ & =2\int (t^3-t+{\displaystyle \frac{t}{1+t^2}})dt\\ & ={\displaystyle \frac{t^4}{2}}-t^2+\ln (1+t^2)+C\\ & ={\displaystyle \frac{{\tan }^4{\displaystyle \frac{x}{2}}}{2}}-{\tan }^2{\displaystyle \frac{x}{2}}+\ln |\sec {\displaystyle \frac{x}{2}}|+C\end{array}$$$$\begin{array}{rl}\text{6.}\int \frac{dx}{\sqrt{x}-\sqrt[3]{x}}& =\int \frac{1}{x^{1/2}-x^{1/3}}dx\\ & =\left|\begin{array}{ll}t=x^{1/6}⟹x=t^6,& dx=6t^{5}dt\\ \sqrt{x}=t^3,& \sqrt[3]{x}=t^2\end{array}\right|\\ & =\int \frac{6t^5}{t^3-t^2}dt\\ & =6\int \frac{t^5}{t^2(t-1)}dt\\ & =6\int \frac{t^3}{t-1}dt\\ & =6\int (t^2+t+1+\frac{1}{t-1})dt\\ & =6(\frac{t^3}{3}+\frac{t^2}{2}+t+\ln |t-1|)+C\\ & =2t^3+3t^2+6t+6\ln |t-1|+C\\ & =2\sqrt{x}+3\sqrt[3]{x}+6x^{\frac{1}{6}}+6\ln |x^{\frac{1}{6}}-1|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{2x^2+4}{x^3+2x^2}dx& =\int ({\displaystyle \frac{2}{x}}+{\displaystyle \frac{4}{x^2}}-{\displaystyle \frac{2}{x+2}})dx\\ & =2\ln |x|-{\displaystyle \frac{4}{x}}-2\ln |x+2|+C\end{array}$$$$\begin{array}{rl}\text{8. }\int \frac{x^3}{(1+x{)}^{10}}dx& =\left|\begin{array}{c}u=1+x⟹x=u-1\\ dx=du\end{array}\right|\\ & =\int {\displaystyle \frac{(u-1{)}^3}{u^{10}}}du\\ & =\int (u^{-7}-3u^{-8}+3u^{-9}-u^{-10})du\\ & =-{\displaystyle \frac{1}{6u^6}}+{\displaystyle \frac{3}{7u^7}}-{\displaystyle \frac{3}{8u^8}}+{\displaystyle \frac{1}{9u^9}}+C\\ & =-{\displaystyle \frac{1}{6(1+x{)}^6}}+{\displaystyle \frac{3}{7(1+x{)}^7}}-{\displaystyle \frac{3}{8(1+x{)}^8}}+{\displaystyle \frac{1}{9(1+x{)}^9}}+C\end{array}$$

变体21

$$\begin{array}{rl}\text{1. }\int \frac{e^x}{1+e^{2x}}dx& =\left|\begin{array}{c}u=e^x\\ du=e^{x}dx\end{array}\right|\\ & =\int \frac{1}{1+u^2}du\\ & =\arctan u+C\\ & =\arctan (e^x)+C\end{array}$$$$\begin{array}{rl}\text{2.}\int x^2\sqrt[5]{1+x^3}dx& =\int x^2(1+x^3{)}^{\frac{1}{5}}dx\\ & =\frac{1}{3}\int (1+x^3{)}^{\frac{1}{5}}\cdot 3x^{2}dx\text{（令 }u=1+x^3⟹du=3x^{2}dx\text{）}\\ & =\frac{1}{3}\int u^{\frac{1}{5}}du\\ & =\frac{1}{3}\cdot \frac{u^{\frac{6}{5}}}{\frac{6}{5}}+C\\ & =\frac{5}{18}{u}^{\frac{6}{5}}+C\\ & =\frac{5}{18}(1+x^3{)}^{\frac{6}{5}}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int e^{\sqrt{2x}}dx& =\left|\begin{array}{c}t=\sqrt{2x}⟹x={\displaystyle \frac{t^2}{2}}\\ dx=tdt\end{array}\right|\\ & =\int e^t\cdot tdt\text{(分部积分)}\\ & =t{e}^t-\int e^{t}dt\\ & =t{e}^t-e^t+C\\ & =\sqrt{2x}{e}^{\sqrt{2x}}-e^{\sqrt{2x}}+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{2x-8}{\sqrt{1-x-x^2}}dx& =\left|\begin{array}{c}\text{配方：}1-x-x^2={\displaystyle \frac{5}{4}}-{(x+{\displaystyle \frac{1}{2}})}^2\\ u=x+{\displaystyle \frac{1}{2}}⟹du=dx\\ \text{分子化简为 }2u-9\end{array}\right|\\ & =-2\sqrt{1-x-x^2}-9\arcsin \left({\displaystyle \frac{2x+1}{\sqrt{5}}}\right)+C\end{array}$$$$\begin{array}{rl}\text{5.}\int {\tan }^{\frac{3}{2}}x{\sec }^{4}xdx& =\int {\tan }^{\frac{3}{2}}x\cdot {\sec }^{2}x\cdot (1+{\tan }^{2}x)dx\\ & =\int {\tan }^{\frac{3}{2}}x\cdot (1+{\tan }^{2}x)\cdot {\sec }^{2}xdx\\ & =\int u^{\frac{3}{2}}(1+u^2)du\text{（令 }u=\tan x⟹du={\sec }^{2}xdx\text{）}\\ & =\int (u^{\frac{3}{2}}+u^{\frac{7}{2}})du\\ & =\frac{2}{5}{u}^{\frac{5}{2}}+\frac{2}{9}{u}^{\frac{9}{2}}+C\\ & =\frac{2}{5}{\tan }^{\frac{5}{2}}x+\frac{2}{9}{\tan }^{\frac{9}{2}}x+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{1+3\cos x}& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\\ \cos x={\displaystyle \frac{1-t^2}{1+t^2}}\end{array}\right|\\ & =\int {\displaystyle \frac{2dt}{4-2t^2}}\\ & ={\displaystyle \frac{1}{\sqrt{2}}}\ln \left|{\displaystyle \frac{\sqrt{2}+t}{\sqrt{2}-t}}\right|+C\\ & ={\displaystyle \frac{1}{\sqrt{2}}}\ln \left|{\displaystyle \frac{\sqrt{2}+\tan {\displaystyle \frac{x}{2}}}{\sqrt{2}-\tan {\displaystyle \frac{x}{2}}}}\right|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{2x^2-3x-3}{(x-1)(x^2-2x+5)}dx& =\int ({\displaystyle \frac{1}{x-1}}+{\displaystyle \frac{x-2}{x^2-2x+5}})dx\\ & =\ln |x-1|+{\displaystyle \frac{1}{2}}\ln (x^2-2x+5)+C\end{array}$$$$\begin{array}{rl}\text{8. }\int \frac{x^5}{\sqrt{x^2-1}}dx& =\left|\begin{array}{c}u=x^2-1⟹x^2=u+1\\ du=2xdx⟹xdx={\displaystyle \frac{1}{2}}du\\ x^5=x\cdot (u+1{)}^2\end{array}\right|\\ & ={\displaystyle \frac{1}{2}}\int (u+1{)}^2{u}^{-1/2}du\\ & ={\displaystyle \frac{1}{2}}({\displaystyle \frac{2}{5}}{u}^{5/2}+{\displaystyle \frac{4}{3}}{u}^{3/2}+2u^{1/2})+C\\ & ={\displaystyle \frac{(x^2-1{)}^{5/2}}{5}}+{\displaystyle \frac{2(x^2-1{)}^{3/2}}{3}}+\sqrt{x^2-1}+C\end{array}$$

变体22

$$\begin{array}{rl}\text{1. }\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx& =\left|\begin{array}{c}u=\sqrt{x}⟹du={\displaystyle \frac{1}{2\sqrt{x}}}dx⟹dx=2udu\end{array}\right|\\ & =2\int e^{u}du\\ & =2e^u+C\\ & =2e^{\sqrt{x}}+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{\sin x}{\cos x+1}dx& =\left|\begin{array}{c}\cos x+1=2{\cos }^2{\displaystyle \frac{x}{2}},\sin x=2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}\end{array}\right|\\ & =\int {\displaystyle \frac{2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}}{2{\cos }^2{\displaystyle \frac{x}{2}}}}dx\\ & =\int \tan {\displaystyle \frac{x}{2}}dx\\ & =-2\ln |\cos {\displaystyle \frac{x}{2}}|+C\\ & =-\ln |\cos x+1|+C\end{array}$$$$\begin{array}{rl}\text{3. }\int \frac{x\cos x}{{\sin }^{3}x}dx& =\left|\begin{array}{c}u=x,dv={\displaystyle \frac{\cos x}{{\sin }^{3}x}}dx\\ du=dx,v=-{\displaystyle \frac{1}{2{\sin }^{2}x}}\end{array}\right|\\ & =-{\displaystyle \frac{x}{2{\sin }^{2}x}}+{\displaystyle \frac{1}{2}}\int {\displaystyle \frac{1}{{\sin }^{2}x}}dx\\ & =-{\displaystyle \frac{x}{2{\sin }^{2}x}}-{\displaystyle \frac{1}{2}}\cot x+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{dx}{x\sqrt{x^2-5x-1}}& =\left|\begin{array}{c}\text{配方：}{x}^2-5x-1={(x-{\displaystyle \frac{5}{2}})}^2-{\displaystyle \frac{29}{4}}\\ u=x-{\displaystyle \frac{5}{2}}⟹du=dx\\ x=u+{\displaystyle \frac{5}{2}}\end{array}\right|\\ & ={\displaystyle \frac{1}{\sqrt{29}}}\ln \left|{\displaystyle \frac{\sqrt{x^2-5x-1}+x-{\displaystyle \frac{5}{2}}}{x}}\right|+C\end{array}$$$$\begin{array}{rl}\text{5. }\int {\cot }^{4}xdx& =\int {\cot }^{2}x({\csc }^{2}x-1)dx\\ & =-{\displaystyle \frac{{\cot }^{3}x}{3}}+\cot x+x+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{1+\sqrt[4]{x}}& =\left|\begin{array}{c}t=\sqrt[4]{x}⟹x=t^4,dx=4t^{3}dt\end{array}\right|\\ & =4\int {\displaystyle \frac{t^3}{1+t}}dt\\ & =4({\displaystyle \frac{t^3}{3}}-{\displaystyle \frac{t^2}{2}}+t-\ln |t+1|)+C\\ & ={\displaystyle \frac{4}{3}}{x}^{\frac{3}{4}}-2\sqrt{x}+4x^{\frac{1}{4}}-4\ln |x^{\frac{1}{4}}+1|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x^2-3x+2}{x(x^2+2x+1)}dx& =\int ({\displaystyle \frac{2}{x}}-{\displaystyle \frac{1}{x+1}}-{\displaystyle \frac{6}{(x+1{)}^2}})dx\\ & =2\ln |x|-\ln |x+1|+{\displaystyle \frac{6}{x+1}}+C\end{array}$$$$\begin{array}{rl}\text{8. }\int \frac{(x-1{)}^2}{(x+5{)}^{10}}dx& =\left|\begin{array}{c}u=x+5⟹x=u-5,dx=du\end{array}\right|\\ & =\int {\displaystyle \frac{(u-6{)}^2}{u^{10}}}du\\ & =\int (u^{-8}-12u^{-9}+36u^{-10})du\\ & =-{\displaystyle \frac{1}{7u^7}}+{\displaystyle \frac{12}{8u^8}}-{\displaystyle \frac{36}{9u^9}}+C\\ & =-{\displaystyle \frac{1}{7(x+5{)}^7}}+{\displaystyle \frac{3}{2(x+5{)}^8}}-{\displaystyle \frac{4}{(x+5{)}^9}}+C\end{array}$$

变体23

$$\begin{array}{rl}\text{1. }\int \frac{e^x}{\sqrt{1-e^{2x}}}dx& =\left|\begin{array}{c}u=e^x\\ du=e^{x}dx⟹dx={\displaystyle \frac{du}{u}}\end{array}\right|\\ & =\int \frac{1}{\sqrt{1-u^2}}du\\ & =\arcsin u+C\\ & =\arcsin (e^x)+C\end{array}$$$$\begin{array}{rl}\text{2. }\int x^4\sqrt{1-6x^5}dx& =\left|\begin{array}{c}u=1-6x^5\\ du=-30x^{4}dx⟹x^{4}dx=-{\displaystyle \frac{1}{30}}du\end{array}\right|\\ & =-{\displaystyle \frac{1}{30}}\int \sqrt{u}du\\ & =-{\displaystyle \frac{1}{30}}\cdot {\displaystyle \frac{2}{3}}{u}^{3/2}+C\\ & =-{\displaystyle \frac{1}{45}}(1-6x^5{)}^{3/2}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int \arctan \sqrt{x}dx& =\left|\begin{array}{c}u=\arctan \sqrt{x},dv=dx\\ du={\displaystyle \frac{1}{1+x}}\cdot {\displaystyle \frac{1}{2\sqrt{x}}}dx,v=x\end{array}\right|\\ & =x\arctan \sqrt{x}-{\displaystyle \frac{1}{2}}\int {\displaystyle \frac{x}{(1+x)\sqrt{x}}}dx\\ & =x\arctan \sqrt{x}-\sqrt{x}+\arctan \sqrt{x}+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{x-1}{\sqrt{3+4x+x^2}}dx& =\left|\begin{array}{c}\text{配方：}{x}^2+4x+3=(x+2{)}^2-1\\ u=x+2⟹du=dx\\ \text{分子化简为 }u-3\end{array}\right|\\ & =\sqrt{(x+2{)}^2-1}-3\ln |x+2+\sqrt{x^2+4x+3}|+C\\ & =\sqrt{x^2+4x+3}-3\ln |x+2+\sqrt{x^2+4x+3}|+C\end{array}$$$$\begin{array}{rl}\text{5. }\int \frac{dx}{\sqrt{\cos x{\sin }^{3}x}}& =\int \frac{dx}{{\sin }^{3/2}x{\cos }^{1/2}x}\\ & =\left|\begin{array}{c}t=\tan x⟹dx={\displaystyle \frac{dt}{1+t^2}}\end{array}\right|\\ & =\int {\displaystyle \frac{(1+t^2{)}^{1/4}}{t^{3/2}}}\cdot {\displaystyle \frac{dt}{1+t^2}}\\ & ={\displaystyle \frac{2}{\sqrt{\sin x}}}+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{2\cos x+3\sin x+5}& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\\ \cos x={\displaystyle \frac{1-t^2}{1+t^2}},\sin x={\displaystyle \frac{2t}{1+t^2}}\end{array}\right|\\ & =\int {\displaystyle \frac{2dt}{(1+t^2)(2\cdot {\displaystyle \frac{1-t^2}{1+t^2}}+3\cdot {\displaystyle \frac{2t}{1+t^2}}+5)}}\\ & ={\displaystyle \frac{1}{\sqrt{21}}}\ln \left|{\displaystyle \frac{5\tan {\displaystyle \frac{x}{2}}-2-\sqrt{21}}{5\tan {\displaystyle \frac{x}{2}}-2+\sqrt{21}}}\right|+C\end{array}$$$$\begin{array}{rl}\text{7.}\int \frac{x+1}{x^4+x^2}dx& =\int \frac{x+1}{x^2(x^2+1)}dx\\ & =\int (\frac{1}{x}+\frac{1}{x^2}-\frac{x+1}{x^2+1})dx\\ & =\int \frac{1}{x}dx+\int \frac{1}{x^2}dx-\int \frac{x}{x^2+1}dx-\int \frac{1}{x^2+1}dx\\ & =\ln |x|-\frac{1}{x}-\frac{1}{2}\ln (x^2+1)-\arctan (x)+C\end{array}$$$$\begin{array}{rl}\text{8.}\int \frac{x+3}{\sqrt{x}+\sqrt{3-x}}dx& =\int \frac{(x+3)(\sqrt{x}-\sqrt{3-x})}{(\sqrt{x}+\sqrt{3-x})(\sqrt{x}-\sqrt{3-x})}dx\\ & =\int \frac{(x+3)(\sqrt{x}-\sqrt{3-x})}{x-(3-x)}dx\\ & =\int \frac{(x+3)(\sqrt{x}-\sqrt{3-x})}{2x-3}dx\\ & =\int \frac{(x+3)\sqrt{x}}{2x-3}dx-\int \frac{(x+3)\sqrt{3-x}}{2x-3}dx\\ & I_1=\left|\begin{array}{c}t=\sqrt{x}⟹x=t^2\\ dx=2tdt\end{array}\right|=\int \frac{(t^2+3)\cdot t}{2t^2-3}\cdot 2tdt=2\int \frac{t^2(t^2+3)}{2t^2-3}dt\\ & =\frac{x^{\frac{3}{2}}}{3}+\frac{9}{2}\sqrt{x}+\frac{9\sqrt{6}}{4}\ln \left|\frac{\sqrt{2x}-\sqrt{3}}{\sqrt{2x}+\sqrt{3}}\right|+C\\ & I_2=\left|\begin{array}{c}t=\sqrt{3-x}⟹x=3-t^2\\ dx=-2tdt\end{array}\right|=-2\int \frac{t^2(6-t^2)}{3-2t^2}dt\\ & =-\frac{(3-x{)}^{\frac{3}{2}}}{3}+3\sqrt{3-x}-\frac{3\sqrt{6}}{2}\ln \left|\frac{\sqrt{3}+\sqrt{6-2x}}{\sqrt{3}-\sqrt{6-2x}}\right|+C\\ & =I_1-I_2\\ & =\frac{x^{\frac{3}{2}}}{3}+\frac{9}{2}\sqrt{x}+\frac{(3-x{)}^{\frac{3}{2}}}{3}-3\sqrt{3-x}+\frac{9\sqrt{6}}{4}\ln \left|\frac{\sqrt{2x}-\sqrt{3}}{\sqrt{2x}+\sqrt{3}}\right|+\frac{3\sqrt{6}}{2}\ln \left|\frac{\sqrt{3}+\sqrt{6-2x}}{\sqrt{3}-\sqrt{6-2x}}\right|+C\\ & =\frac{x^{\frac{3}{2}}}{3}+\frac{(3-x{)}^{\frac{3}{2}}}{3}+\frac{9}{2}\sqrt{x}-3\sqrt{3-x}+\frac{3\sqrt{6}}{4}(3\ln \left|\frac{\sqrt{2x}-\sqrt{3}}{\sqrt{2x}+\sqrt{3}}\right|+2\ln \left|\frac{\sqrt{3}+\sqrt{6-2x}}{\sqrt{3}-\sqrt{6-2x}}\right|)+C\end{array}$$

变体24

$$\begin{array}{rl}\text{1. }\int \frac{\ln x}{x(1+{\ln }^{4}x)}dx& =\left|\begin{array}{c}u=\ln x\\ du={\displaystyle \frac{1}{x}}dx\end{array}\right|\\ & =\int \frac{u}{1+u^4}du\\ & =\left|\begin{array}{c}t=u^2\\ dt=2udu\end{array}\right|\\ & ={\displaystyle \frac{1}{2}}\int \frac{1}{1+t^2}dt\\ & ={\displaystyle \frac{1}{2}}\arctan t+C\\ & ={\displaystyle \frac{1}{2}}\arctan ({\ln }^{2}x)+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{\sin \sqrt{x-1}}{\sqrt{x-1}}dx& =\left|\begin{array}{c}u=\sqrt{x-1}\\ du={\displaystyle \frac{1}{2\sqrt{x-1}}}dx⟹dx=2udu\end{array}\right|\\ & =2\int \sin udu\\ & =-2\cos u+C\\ & =-2\cos \sqrt{x-1}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int (x^2-1)e^{x}dx& =\left|\begin{array}{c}u=x^2-1,dv=e^{x}dx\\ du=2xdx,v=e^x\end{array}\right|\\ & =(x^2-1)e^x-2\int x{e}^{x}dx\\ & =(x^2-1)e^x-2(x{e}^x-\int e^{x}dx)\\ & =(x^2-1)e^x-2x{e}^x+2e^x+C\\ & =(x^2-2x+1)e^x+C\\ & =(x-1{)}^2{e}^x+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{2x-3}{\sqrt{x^2+4x+5}}dx& =\left|\begin{array}{c}\text{配方：}{x}^2+4x+5=(x+2{)}^2+1\\ u=x+2⟹du=dx\\ \text{分子化简为 }2u-7\end{array}\right|\\ & =2\int \frac{u}{\sqrt{u^2+1}}du-7\int \frac{1}{\sqrt{u^2+1}}du\\ & =2\sqrt{u^2+1}-7\ln |u+\sqrt{u^2+1}|+C\\ & =2\sqrt{x^2+4x+5}-7\ln |x+2+\sqrt{x^2+4x+5}|+C\end{array}$$$$\begin{array}{rl}\text{5. }\int \frac{\cos 2x}{{\sin }^{4}x}dx& =\int \frac{1-2{\sin }^{2}x}{{\sin }^{4}x}dx\\ & =\int {\csc }^{4}xdx-2\int {\csc }^{2}xdx\\ & =-\cot x-{\displaystyle \frac{{\cot }^{3}x}{3}}+2\cot x+C\\ & =\cot x-{\displaystyle \frac{{\cot }^{3}x}{3}}+C\end{array}$$$$\begin{array}{rl}\text{6.}\int \frac{dx}{\sqrt[3]{x}-\sqrt[6]{x}}& =\int \frac{dx}{x^{1/3}-x^{1/6}}\\ & =\left|\begin{array}{c}t=x^{1/6}⟹x=t^6\Rightarrow dx=6t^{5}dt\\ x^{1/3}=t^2,x^{1/6}=t\end{array}\right|\\ & =\int \frac{6t^5}{t^2-t}dt\\ & =6\int \frac{t^5}{t(t-1)}dt\\ & =6\int \frac{t^4}{t-1}dt\\ & =6\int (t^3+t^2+t+1+\frac{1}{t-1})dt\\ & =6(\frac{t^4}{4}+\frac{t^3}{3}+\frac{t^2}{2}+t+\ln |t-1|)+C\\ & =\frac{3}{2}{t}^4+2t^3+3t^2+6t+6\ln |t-1|+C\\ & =\frac{3}{2}{x}^{2/3}+2x^{1/2}+3x^{1/3}+6x^{1/6}+6\ln |x^{1/6}-1|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{xdx}{x^3+1}& =\int (-{\displaystyle \frac{1}{3(x+1)}}+{\displaystyle \frac{x+1}{3(x^2-x+1)}})dx\\ & =-{\displaystyle \frac{1}{3}}\ln |x+1|+{\displaystyle \frac{1}{6}}\ln (x^2-x+1)+{\displaystyle \frac{\sqrt{3}}{3}}\arctan \left({\displaystyle \frac{2x-1}{\sqrt{3}}}\right)+C\end{array}$$$$\begin{array}{rl}\text{8. }\int \frac{x^3}{\sqrt{9-x^2}}dx& =\left|\begin{array}{c}u=9-x^2\\ du=-2xdx⟹xdx=-{\displaystyle \frac{1}{2}}du\\ x^3=x\cdot (9-u)\end{array}\right|\\ & =-{\displaystyle \frac{1}{2}}\int {\displaystyle \frac{9-u}{\sqrt{u}}}du\\ & =-{\displaystyle \frac{1}{2}}(18\sqrt{u}-{\displaystyle \frac{2}{3}}{u}^{3/2})+C\\ & =-9\sqrt{9-x^2}+{\displaystyle \frac{1}{3}}(9-x^2{)}^{3/2}+C\end{array}$$

变体25

$$\begin{array}{rl}\text{1. }\int e^x\sqrt{3-2e^x}dx& =\left|\begin{array}{c}u=3-2e^x\\ du=-2e^{x}dx⟹e^{x}dx=-{\displaystyle \frac{1}{2}}du\end{array}\right|\\ & =-{\displaystyle \frac{1}{2}}\int \sqrt{u}du\\ & =-{\displaystyle \frac{1}{3}}{u}^{\frac{3}{2}}+C\\ & =-{\displaystyle \frac{1}{3}}(3-2e^x{)}^{\frac{3}{2}}+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{1}{x^2}{e}^{\frac{1}{x}}dx& =\left|\begin{array}{c}u={\displaystyle \frac{1}{x}}\\ du=-{\displaystyle \frac{1}{x^2}}dx⟹dx=-x^{2}du\end{array}\right|\\ & =-\int e^{u}du\\ & =-e^u+C\\ & =-e^{\frac{1}{x}}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int \cos \sqrt{x}dx& =\left|\begin{array}{c}t=\sqrt{x}⟹x=t^2\\ dx=2tdt\end{array}\right|\\ & =2\int t\cos tdt\text{(分部积分)}\\ & =2(t\sin t-\int \sin tdt)\\ & =2(t\sin t+\cos t)+C\\ & =2\sqrt{x}\sin \sqrt{x}+2\cos \sqrt{x}+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{dx}{x\sqrt{x^2+6x+2}}& =\int \frac{dx}{x\sqrt{(x+3{)}^2-7}}\\ & =\left|\begin{array}{c}u=x+3⟹x=u-3\\ du=dx\end{array}\right|\\ & ={\displaystyle \frac{1}{\sqrt{7}}}\ln \left|{\displaystyle \frac{\sqrt{x^2+6x+2}+x+3}{x}}\right|+C\end{array}$$$$\begin{array}{rl}\text{5. }\int {\tan }^{4}2xdx& =\int ({\sec }^{2}2x-1{)}^{2}dx\\ & ={\displaystyle \frac{{\tan }^{3}2x}{6}}-{\displaystyle \frac{\tan 2x}{2}}+x+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{3\sin x-4\cos x-3}& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\\ \sin x={\displaystyle \frac{2t}{1+t^2}},\cos x={\displaystyle \frac{1-t^2}{1+t^2}}\end{array}\right|\\ & ={\displaystyle \frac{1}{4}}\ln \left|{\displaystyle \frac{\tan {\displaystyle \frac{x}{2}}-1}{\tan {\displaystyle \frac{x}{2}}+7}}\right|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x^3+2}{x^3+4x}dx& =\int (1-{\displaystyle \frac{4x-2}{x(x^2+4)}})dx\\ & =x+{\displaystyle \frac{1}{2}}\ln |x|-{\displaystyle \frac{9}{4}}\ln (x^2+4)+C\end{array}$$$$\begin{array}{rl}\text{8. }\int (3x+1{)}^2\sqrt{x-1}dx& =\int (9x^2+6x+1)\sqrt{x-1}dx\\ & ={\displaystyle \frac{18}{7}}(x-1{)}^{\frac{7}{2}}+{\displaystyle \frac{48}{5}}(x-1{)}^{\frac{5}{2}}+14(x-1{)}^{\frac{3}{2}}+8\sqrt{x-1}+C\end{array}$$

变体26

$$\begin{array}{rl}\text{1. }\int \frac{\sqrt[5]{\arcsin x}}{\sqrt{1-x^2}}dx& =\left|\begin{array}{c}u=\arcsin x\\ du={\displaystyle \frac{1}{\sqrt{1-x^2}}}dx⟹dx=\sqrt{1-x^2}du\end{array}\right|\\ & =\int u^{1/5}du\\ & ={\displaystyle \frac{u^{6/5}}{6/5}}+C\\ & ={\displaystyle \frac{5}{6}}{u}^{6/5}+C\\ & ={\displaystyle \frac{5}{6}}(\arcsin x{)}^{6/5}+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{1-\sin 5x}{{\cos }^{2}5x}dx& =\int {\sec }^{2}5xdx-\int \tan 5x\sec 5xdx\\ & =\frac{1}{5}\tan 5x-\frac{1}{5}\sec 5x+C\end{array}$$$$\begin{array}{rl}\text{3. }\int \arcsin xdx& =\left|\begin{array}{c}u=\arcsin x,dv=dx\\ du={\displaystyle \frac{1}{\sqrt{1-x^2}}}dx,v=x\end{array}\right|\\ & =x\arcsin x-\int \frac{x}{\sqrt{1-x^2}}dx\\ & =x\arcsin x+\sqrt{1-x^2}+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{1+x}{\sqrt{6x-x^2}}dx& =\int \frac{(x-3)+4}{\sqrt{9-(x-3{)}^2}}dx\\ & =-\sqrt{6x-x^2}+4\arcsin \left(\frac{x-3}{3}\right)+C\end{array}$$$$\begin{array}{rl}\text{5. }\int {\cos }^{3}2xdx& =\int \cos 2x(1-{\sin }^{2}2x)dx\\ & =\frac{1}{2}\sin 2x-\frac{1}{6}{\sin }^{3}2x+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{1-\sqrt{x}}& =\left|\begin{array}{c}t=\sqrt{x}⟹x=t^2,dx=2tdt\end{array}\right|\\ & =-2t-2\ln |t-1|+C\\ & =-2\sqrt{x}-2\ln |\sqrt{x}-1|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x^2+9x+9}{x^2(x^2+9)}dx& =\int (\frac{1}{x}+\frac{1}{x^2}-\frac{x}{x^2+9})dx\\ & =\ln |x|-\frac{1}{x}-\frac{1}{2}\ln (x^2+9)+C\end{array}$$$$\begin{array}{rl}\text{8. }\int x^3\sqrt{(4+x^2{)}^{13}}dx& =\int x^3(4+x^2{)}^{\frac{13}{2}}dx\\ & =\left|\begin{array}{c}u=4+x^2⟹du=2xdx\\ x^3=x\cdot (u-4)\end{array}\right|\\ & =\frac{1}{2}\int (u-4)u^{\frac{13}{2}}du\\ & =\frac{(4+x^2{)}^{\frac{17}{2}}}{17}-\frac{4(4+x^2{)}^{\frac{15}{2}}}{15}+C\end{array}$$

变体27

$$\begin{array}{rl}\text{1. }\int \frac{\ln x\sqrt{1+{\ln }^{2}x}}{x}dx& =\left|\begin{array}{c}u=1+{\ln }^{2}x\\ du=2\ln x\cdot {\displaystyle \frac{1}{x}}dx⟹\ln x\cdot {\displaystyle \frac{1}{x}}dx={\displaystyle \frac{1}{2}}du\end{array}\right|\\ & ={\displaystyle \frac{1}{2}}\int \sqrt{u}du\\ & ={\displaystyle \frac{1}{3}}{u}^{\frac{3}{2}}+C\\ & ={\displaystyle \frac{1}{3}}(1+{\ln }^{2}x{)}^{\frac{3}{2}}+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{e^{2/x}}{x^2}dx& =\left|\begin{array}{c}u=-{\displaystyle \frac{2}{x}}\\ du={\displaystyle \frac{2}{x^2}}dx⟹{\displaystyle \frac{1}{x^2}}dx=-{\displaystyle \frac{1}{2}}du\end{array}\right|\\ & =-{\displaystyle \frac{1}{2}}\int e^{u}du\\ & =-{\displaystyle \frac{1}{2}}{e}^u+C\\ & =-{\displaystyle \frac{1}{2}}{e}^{2/x}+C\end{array}$$$$\begin{array}{rl}\text{3. }\int x^3{e}^{2x^2}dx& =\left|\begin{array}{c}u=x^2\\ du=2xdx⟹xdx={\displaystyle \frac{1}{2}}du\\ x^3=u\cdot x⟹x^{3}dx={\displaystyle \frac{1}{2}}u{e}^{2u}du\end{array}\right|\\ & ={\displaystyle \frac{1}{2}}\int u{e}^{2u}du\text{(分部积分)}\\ & ={\displaystyle \frac{1}{4}}u{e}^{2u}-{\displaystyle \frac{1}{8}}{e}^{2u}+C\\ & ={\displaystyle \frac{1}{4}}{x}^2{e}^{2x^2}-{\displaystyle \frac{1}{8}}{e}^{2x^2}+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{(3x+7)dx}{\sqrt{1+6x-x^2}}& =\left|\begin{array}{c}\text{配方：}1+6x-x^2=10-(x-3{)}^2\\ u=x-3⟹du=dx\\ \text{分子化简为 }3u+16\end{array}\right|\\ & =\int \frac{3u+16}{\sqrt{10-u^2}}du\\ & =-3\sqrt{10-u^2}+16\arcsin \left({\displaystyle \frac{u}{\sqrt{10}}}\right)+C\\ & =-3\sqrt{1+6x-x^2}+16\arcsin \left({\displaystyle \frac{x-3}{\sqrt{10}}}\right)+C\end{array}$$$$\begin{array}{rl}\text{5. }\int \frac{1+\tan x}{\sin 2x}dx& =\int \frac{1+\tan x}{2\sin x\cos x}dx\\ & ={\displaystyle \frac{1}{2}}(\int \frac{1}{\sin x\cos x}dx+\int \frac{\tan x}{\sin x\cos x}dx)\\ & ={\displaystyle \frac{1}{2}}(\ln |\tan x|+\tan x)+C\end{array}$$$$\begin{array}{rl}\text{6. }\int \frac{dx}{2\sin x+5\cos x+1}& =\left|\begin{array}{c}t=\tan {\displaystyle \frac{x}{2}}⟹dx={\displaystyle \frac{2}{1+t^2}}dt\\ \sin x={\displaystyle \frac{2t}{1+t^2}},\cos x={\displaystyle \frac{1-t^2}{1+t^2}}\end{array}\right|\\ & ={\displaystyle \frac{1}{\sqrt{21}}}\ln \left|{\displaystyle \frac{5\tan {\displaystyle \frac{x}{2}}-2-\sqrt{21}}{5\tan {\displaystyle \frac{x}{2}}-2+\sqrt{21}}}\right|+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x^2}{(x^2+2)(x-4)}dx& =\int ({\displaystyle \frac{8}{9(x-4)}}+{\displaystyle \frac{\frac{1}{9}x+\frac{4}{9}}{x^2+2}})dx\\ & ={\displaystyle \frac{8}{9}}\ln |x-4|+{\displaystyle \frac{1}{18}}\ln (x^2+2)+{\displaystyle \frac{2\sqrt{2}}{9}}\arctan \left({\displaystyle \frac{x}{\sqrt{2}}}\right)+C\end{array}$$$$\begin{array}{rl}\text{8.}\int \frac{x-1}{\sqrt{x}-\sqrt{1+x}}dx& =\int \frac{(x-1)(\sqrt{x}+\sqrt{1+x})}{(\sqrt{x}-\sqrt{1+x})(\sqrt{x}+\sqrt{1+x})}dx\\ & =\int \frac{(x-1)(\sqrt{x}+\sqrt{1+x})}{x-(1+x)}dx\\ & =-\int (x-1)(\sqrt{x}+\sqrt{1+x})dx\\ & =-\int (x\sqrt{x}+x\sqrt{1+x}-\sqrt{x}-\sqrt{1+x})dx\\ & =-\int x^{3/2}dx-\int x\sqrt{1+x}dx+\int \sqrt{x}dx+\int \sqrt{1+x}dx\\ & =-(\frac{2}{5}{x}^{5/2}+\frac{2}{5}(1+x{)}^{5/2}-\frac{2}{3}(1+x{)}^{3/2})+\frac{2}{3}{x}^{3/2}+\frac{2}{3}(1+x{)}^{3/2}+C\\ & =-\frac{2}{5}{x}^{5/2}-\frac{2}{5}(1+x{)}^{5/2}+\frac{4}{3}(1+x{)}^{3/2}+\frac{2}{3}{x}^{3/2}+C\end{array}$$

变体28

$$\begin{array}{rl}\text{1. }\int \frac{e^{-x}}{e^{-2x}-1}dx& =\left|\begin{array}{c}u=e^{-x}\\ du=-e^{-x}dx⟹dx=-{\displaystyle \frac{du}{u}}\end{array}\right|\\ & =-\int \frac{1}{u^2-1}du\\ & =-{\displaystyle \frac{1}{2}}\ln \left|{\displaystyle \frac{u-1}{u+1}}\right|+C\\ & ={\displaystyle \frac{1}{2}}\ln \left|{\displaystyle \frac{1+e^x}{1-e^x}}\right|+C\end{array}$$$$\begin{array}{rl}\text{2. }\int \frac{x^2}{x^6+1}dx& =\left|\begin{array}{c}u=x^3\\ du=3x^{2}dx⟹x^{2}dx={\displaystyle \frac{1}{3}}du\end{array}\right|\\ & ={\displaystyle \frac{1}{3}}\int \frac{1}{u^2+1}du\\ & ={\displaystyle \frac{1}{3}}\arctan (u)+C\\ & ={\displaystyle \frac{1}{3}}\arctan (x^3)+C\end{array}$$$$\begin{array}{rl}\text{3. }\int x^3{\ln }^{2}xdx& =\left|\begin{array}{c}u={\ln }^{2}x,dv=x^{3}dx\\ du={\displaystyle \frac{2\ln x}{x}}dx,v={\displaystyle \frac{x^4}{4}}\end{array}\right|\\ & ={\displaystyle \frac{x^4}{4}}{\ln }^{2}x-{\displaystyle \frac{1}{2}}\int x^3\ln xdx\\ & ={\displaystyle \frac{x^4}{4}}{\ln }^{2}x-{\displaystyle \frac{x^4}{8}}\ln x+{\displaystyle \frac{x^4}{32}}+C\end{array}$$$$\begin{array}{rl}\text{4. }\int \frac{dx}{x\sqrt{3x^2-2x-1}}& =\left|\begin{array}{c}\text{配方：}3x^2-2x-1=3{(x-{\displaystyle \frac{1}{3}})}^2-{\displaystyle \frac{4}{3}}\\ u=x-{\displaystyle \frac{1}{3}}⟹du=dx\\ x=u+{\displaystyle \frac{1}{3}}\end{array}\right|\\ & ={\displaystyle \frac{1}{\sqrt{3}}}\ln \left|{\displaystyle \frac{\sqrt{3x^2-2x-1}+\sqrt{3}x-1}{x}}\right|+C\end{array}$$$$\begin{array}{rl}\text{5. }\int {(\sin x+\frac{1}{\cos x})}^{2}dx& =\int ({\sin }^{2}x+2\tan x+{\sec }^{2}x)dx\\ & ={\displaystyle \frac{x}{2}}-{\displaystyle \frac{\sin 2x}{4}}+2\ln |\sec x|+\tan x+C\end{array}$$$$\begin{array}{rl}\text{6.}\int \frac{\sqrt[4]{x+3}}{\sqrt{x+3}+4}dx& =\left|\begin{array}{c}t=\sqrt{x+3}⟹x+3=t^2⟹dx=2tdt\\ \sqrt[4]{x+3}=t^{1/2}\end{array}\right|\\ & =\int \frac{t^{1/2}}{t+4}\cdot 2tdt\\ & =2\int \frac{t^{3/2}}{t+4}dt\\ & =\left|\begin{array}{c}u=t^{1/2}⟹t=u^2⟹dt=2udu\\ t^{3/2}=u^3\end{array}\right|\\ & =4\int \frac{u^4}{u^2+4}du\\ & =4\int (u^2-4+\frac{16}{u^2+4})du\\ & =4(\frac{u^3}{3}-4u+8\arctan \left(\frac{u}{2}\right))+C\\ & =\frac{4u^3}{3}-16u+32\arctan \left(\frac{u}{2}\right)+C\\ & =\frac{4}{3}(x+3{)}^{3/4}-16(x+3{)}^{1/4}+32\arctan \left(\frac{(x+3{)}^{1/4}}{2}\right)+C\end{array}$$$$\begin{array}{rl}\text{7. }\int \frac{x^2+x-3}{x^3(x-1)}dx& =\int ({\displaystyle \frac{3}{x}}-{\displaystyle \frac{1}{x^2}}+{\displaystyle \frac{1}{x-1}})dx\\ & =3\ln |x|+{\displaystyle \frac{1}{x}}+\ln |x-1|+C\end{array}$$$$\begin{array}{rl}\text{8. }\int (x+1)(3x-1{)}^{7}dx& =\left|\begin{array}{c}u=3x-1⟹x={\displaystyle \frac{u+1}{3}}\\ dx={\displaystyle \frac{1}{3}}du\end{array}\right|\\ & ={\displaystyle \frac{1}{27}}\int (u+4)u^{7}du\\ & ={\displaystyle \frac{1}{27}}({\displaystyle \frac{u^9}{9}}+{\displaystyle \frac{4u^8}{2}})+C\\ & ={\displaystyle \frac{(3x-1{)}^9}{243}}+{\displaystyle \frac{2(3x-1{)}^8}{27}}+C\end{array}$$