答案：

$$\begin{array}{rlrl}& \int (\sqrt{x\sqrt{x\sqrt{x}}}+3^x{e}^x)dx& & \sqrt{x\sqrt{x\sqrt{x}}}=(x\cdot (x\cdot (x{)}^{\frac{1}{2}}{)}^{\frac{1}{2}}{)}^{\frac{1}{2}}\\ & =\int (x^{\frac{7}{8}}+3^x{e}^x)dx& & =x^{\frac{1}{x}}\cdot (x\cdot (x{)}^{\frac{1}{2}}{)}^{\frac{1}{4}}\\ & =\int x^{\frac{7}{8}}dx+\int 3^x{e}^{x}dx& & =x^{\frac{1}{x}}\cdot x^{\frac{1}{4}}\cdot x^{\frac{1}{8}}\\ & =\frac{8}{15}{x}^{\frac{15}{8}}+\int (3e{)}^{x}dx& & =x^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}\\ & =\frac{8}{15}{x}^{\frac{15}{8}}+\frac{{\left(3e\right)}^x}{\ln 3e}+c& & =x^{\frac{4}{8}+\frac{2}{8}+\frac{1}{8}}\\ & & & =x^{\frac{7}{8}}\end{array}$$

答案：

$$\begin{array}{rl}& \int \frac{dx}{\sqrt{3+2x-x^2}}\\ & =\int \frac{1}{\sqrt{4-(x-1{)}^2}}dx\\ & =\int \frac{1}{\sqrt{(2{)}^2-(x-1{)}^2}}d(x-1)\\ & =\arcsin \frac{x-1}{2}+c\end{array}$$

答案：

$$\begin{array}{rl}& \int \tan \sqrt{1+x^2}\cdot \frac{x}{\sqrt{1+x^2}}dx\\ & =\frac{1}{2}\int \tan \sqrt{x^2+1}\cdot \frac{1}{\sqrt{1+x^2}}d(x^2+1)\\ & =\frac{1}{2}\int \tan \sqrt{t}\cdot \frac{1}{\sqrt{t}}dt\\ & =\int \tan \sqrt{t}d\sqrt{t}\\ & =-\ln |\cos \sqrt{t}|+c\\ & =-\ln |\cos \sqrt{x^2+1}|+C\end{array}$$

答案：

$$\begin{array}{rl}& \int \frac{1}{e^x+e^{-x}}dx\\ & =\int \frac{e^x}{e^{2x}+1}dx\\ & =\int \frac{1}{e^{2x}+1}d{e}^x\\ & =\int \frac{1}{{\left(e^x\right)}^2+{\left(1\right)}^2}d{e}^x\\ & =\arctan e^x+C\end{array}$$

答案：

$$\begin{array}{rl}& \int \frac{1}{e^x+1}dx\\ & =\int \frac{e^{-x}}{1+e^{-x}}dx\\ & =\int \frac{1}{1+e^{-x}}d{e}^{-x}\\ & =-\int \frac{1}{1+e^{-x}}d(e^{-x}+1)\\ & =-\ln |e^{-x}+1|+C\\ & =-\ln (e^{-x}+1)+C\end{array}$$

答案：

$$\begin{array}{rl}& \int {\tan }^{3}xdx\\ & =\int \tan x({\sec }^{2}x-1)dx\\ & =\int \tan x{\sec }^{2}xdx-\int \tan xdx\\ & =\int \tan xd\tan x+\ln |\cos x|\\ & =\frac{1}{2}{\tan }^{2}x+\ln |\cos x|+c\end{array}$$

答案：

$$\begin{array}{c}=\int \frac{\frac{1-\ln x}{x^2}}{{(1-\frac{\ln x}{x})}^2}\mathrm{d}x=\int \frac{{\left(\frac{\ln x}{x}\right)}^{\mathrm{\prime }}}{{(1-\frac{\ln x}{x})}^2}\mathrm{d}x=\int \frac{1}{{(1-\frac{\ln x}{x})}^2}\mathrm{d}(\frac{\ln x}{x})\\ =-\int \frac{1}{{(1-\frac{\ln x}{x})}^2}\mathrm{d}(\begin{array}{ccc}1& -\frac{\ln x}{x})& =\frac{1}{1-\frac{\ln x}{x}}+C=\frac{x}{x-\ln x}+C.\end{array}\end{array}$$

答案：

$$\begin{array}{rl}\text{令 }\sin x& =A(\sin x+\cos x{)}^{\mathrm{\prime }}+B(\sin x+\cos x)\\ & \\ & =(A+B)\cos x+(B-A)\sin x,\\ & \\ & \text{则}\{\begin{array}{c}A+B=0,\\ B-A=1\end{array}\Rightarrow A=-\frac{1}{2},B=\frac{1}{2}.\text{故}\\ & \\ & \begin{array}{rl}\text{原式}& =\int \frac{-\frac{1}{2}(\sin x+\cos x{)}^{\mathrm{\prime }}+\frac{1}{2}(\sin x+\cos x)}{\sin x+\cos x}\mathrm{d}x\\ & =-\frac{1}{2}\int \frac{1}{\sin x+\cos x}\mathrm{d}(\sin x+\cos x)+\frac{1}{2}x\\ & =-\frac{1}{2}\mathrm{l}\mathrm{n}|\sin x+\cos x|+\frac{1}{2}x+C.\end{array}\end{array}$$

答案：

$$\begin{array}{rl}& \int \frac{7\cos x-3\sin x}{5\cos x+2\sin x}dx\\ & =\int \frac{(5\cos x+2\sin x{)}^{\mathrm{\prime }}+(5\cos x+2\sin x)}{5\cos x+2\sin x}dx\\ & =\int \frac{(5\cos x+2\sin x{)}^{\mathrm{\prime }}+(5\cos x+2\sin x)}{5\cos x+2\sin x}dx\\ & =\int \frac{(5\cos x+2\sin x{)}^{\mathrm{\prime }}}{5\cos x+2\sin x}dx+\int \frac{5\cos x+2\sin x}{5\cos x+2\sin x}dx\\ & =\ln |5\cos x+2\sin x|+x+C\end{array}$$

答案：

$$\begin{array}{rlrlrl}& \int \frac{x^2}{\sqrt{a^2-x^2}}dx& & 令x=a\sin t& & t=\arcsin \frac{x}{a}\\ & =\int \frac{a^2{\sin }^{2}t}{\sqrt{a^2-a^2{\sin }^{2}t}}\cdot a\cos tdt& & dx=a\cos tdt& & \\ & & & \sin 2t=2\sin t\cos t& & \\ & =\int \frac{a^2{\sin }^{2}t}{a\cos t}\cdot a\cos tdt& & \sin t=\frac{x}{a}& & \cos t=\frac{\sqrt{a^2-x^2}}{a}\\ & =\int a^2{\sin }^{2}tdt& & \sin 2t=\frac{2\times \sqrt{a^2-x^2}}{a^2}& & \\ & =a^2\int {\sin }^{2}tdt& & & & \\ & =a^2\int \frac{1-\cos t}{2}dt& & & & \\ & =\frac{a^2}{2}\int 1-\cos 2tdt& & & & \\ & =\frac{a^2}{2}{\int }_{}1dt-\frac{a^2}{2}\int \cos 2tdt& & & & \\ & =\frac{a^2}{2}t-\frac{a^2}{4}\int \cos 2td2t& & & & \\ & =\frac{a^2}{2}t-\frac{a^2}{4}\sin 2t+c& & & & \\ & =\frac{a^2}{2}\arcsin \frac{x}{a}-\frac{x\sqrt{a^2-x^2}}{2}+C& & & & \end{array}$$

心得：

这道题目能够让你困惑的就是三角换元的回代问题了，sint，cost这些怎么求出来的？

瞧好了，小子，我只演示一遍

答案：

$$\begin{array}{rlrl}& \int \frac{1}{x^2\sqrt{x^2+1}}dx& & \\ & & & \sqrt{x^2+a^2}令x=a\tan t\\ & =\int \frac{{\sec }^{2}t}{{\tan }^{2}t\sqrt{{\tan }^{2}t+1}}dt& & 即x=\tan t\\ & & & dx={\sec }^{2}tdt\\ & =\int \frac{{\sec }^{2}t}{{\tan }^{2}t\cdot \sec t}dt& & t=arc\tan x\\ & =\int \frac{\sec t}{{\tan }^{2}t}dt& & \sin t=\frac{x}{\sqrt{1^2+x^2}}\\ & =\int \frac{1}{\cos t}\cdot \frac{{\cos }^{2}t}{{\sin }^{2}t}dt& & \cos t=\frac{1}{\sqrt{1^2+x^2}}\\ & =\int \frac{\cos t}{{\sin }^{2}t}dt& & \\ & =\int \cos t{\csc }^{2}tdt& & \\ & =-\int \cos td(\cot t)& & \\ & =-\cos t\cdot \cot t-\int \cot t\cdot \sin tdt& & \\ & =-\cos t\cdot \frac{\cos t}{\sin t}-\int \cos tdt& & \\ & & & =-\frac{1}{1+x^2}\cdot \frac{\sqrt{1+x^2}}{x}-\frac{x}{\sqrt{1+x^2}}+c=-\frac{1}{x\sqrt{1+x^2}}-\frac{x}{\sqrt{1+x^2}}=-\frac{1+x^2}{x\sqrt{1+x^2}}+c=-\frac{\sqrt{1+x^2}}{x}+c\\ & & & \end{array}$$

答案：

$$\begin{array}{rl}& \text{令}\sqrt[6]{x}=t,\text{则 }x=t^6.\text{ 故}\\ & \text{原式}=\int \frac{t^3}{t^3-t^2}\cdot 6t^5\mathrm{d}t=6\int \frac{t^6}{t-1}\mathrm{d}t\\ & =6\int \frac{t^5(t-1)+t^4(t-1)+t^3(t-1)+t^2(t-1)+t(t-1)+t-1+1}{t-1}\mathrm{d}t\\ & =6\int (t^5+t^4+t^3+t^2+t+1+\frac{1}{t-1})\mathrm{d}t\\ & =t^6+\frac{6}{5}{t}^5+\frac{3}{2}{t}^4+2t^3+3t^2+6t+6\mathrm{l}\mathrm{n}|t-1|+C\\ & =x+\frac{6}{5}{x}^{\frac{5}{6}}+\frac{3}{2}{x}^{\frac{2}{3}}+2x^{\frac{1}{2}}+3x^{\frac{1}{3}}+6x^{\frac{1}{6}}+6\ln |\sqrt[6]{x}-1|+C.\end{array}$$

答案：

$$\begin{array}{rl}& \text{令}\sqrt{x+1}=t\text{ ,则 }x=t^2-1.\text{ 故}\\ & \text{原式}=\int \frac{1+t}{1-t}\cdot 2t\mathrm{d}t=-2\int \frac{t^2+t}{t-1}\mathrm{d}t=-2\int \frac{(t^2-1)+(t-1)+2}{t-1}\mathrm{d}t\\ & =-2\int (t+2+\frac{2}{t-1})\mathrm{d}t=-t^2-4t-4\ln |t-1|+C\\ & =-(\begin{array}{c}x+1\end{array})-4\sqrt{x+1}-4\ln |\begin{array}{c}\sqrt{x+1}\end{array}-1|+C\\ & =-x-4\sqrt{x+1}-4\ln |\sqrt{x+1}-1|+C_1(\text{ 其中 }{C}_1=C-1).\end{array}$$

答案：

$$\begin{array}{rl}& \int \sqrt{\frac{1-x}{1+x}}\cdot \frac{1}{x}dx\\ & =\int t\cdot \frac{1+t^2}{1-t^2}\cdot \frac{-4t}{(1+t^2{)}^2}dt\\ & =\int \frac{-4t^2}{(1-t^2)(1+t^2)}dt\\ & =-4\int \frac{t^2}{(1-t^2)(1+t^2)}dt\\ & =-2\int \frac{1}{1-t^2}-\frac{1}{1+t^2}dt\\ & =-2\int \frac{1}{1-t^2}dt+2\int \frac{1}{1+t^2}dt\\ & =-2\int \frac{1}{2}(\frac{1}{1+t}+\frac{1}{1-t})dt+2\arctan t\\ & =-\int \frac{1}{1+t}+\frac{1}{1-t}dt+2\arctan t\end{array}\begin{array}{l}=-\ln |1+t|+\ln |1-t|+2\arctan t+c\\ 又\because t=\sqrt{\frac{1-x}{1+x}}\\ =\ln \left|\frac{1-t}{1+t}\right|+2\mathrm{corctan}t+c\\ =\ln \left|\frac{1-\sqrt{\frac{1-x}{1+x}}}{1+\sqrt{\frac{1-x}{1+x}}}\right|+2\arctan \sqrt{\frac{1-x}{1+x}}+c\\ =\ln \left|\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right|+2\arctan \sqrt{\frac{1-x}{1+x}}+c\end{array}$$

答案：

$$\begin{array}{rl}\sqrt{e^x-1}& =t\\ e^x& -1=t^2\\ e^x& =t^2+1\\ x& =\ln (t^2+1)\\ dx& =\frac{2t}{t^2+1}dt\end{array}$$

$$\begin{array}{rl}& \int \frac{x{e}^x}{\sqrt{e^x-1}}dx\\ & =\int \frac{x}{\sqrt{e^x-1}}d{e}^x-1\\ & =2\int xd\sqrt{e^x-1}\\ & =2x\sqrt{e^x-1}-2\int \sqrt{e^x-1}dx\\ & =2x\sqrt{e^x-1}-2\int t\cdot \frac{2t}{t^2+1}dt\\ & =2x\sqrt{e^x-1}-4\int \frac{t^2+1-1}{t^2+1}dt\\ & =2x\sqrt{e^x-1}-4{\int }_{}1dt+4\int \frac{1}{t^2+1}dt\\ & =2x\sqrt{e^x-1}-4t+4\arctan t+c\\ & =2x\sqrt{e^x-1}-4\sqrt{e^x-1}+4arc\tan \sqrt{e^x-1}+c\end{array}$$

答案：

$$\begin{array}{rl}& \int \frac{\ln (1+x)}{\sqrt{x}}dx\\ & 令\sqrt{x}=t即x=t^2\\ & =\int \frac{\ln (1+t^2)}{t}2tdt\\ & =2\int \ln (1+t^2)dt\\ & =2\ln {(1+t^2)}^{}t-2\int t\frac{2t}{1+t^2}dt\\ & =2\ln {(t+t^2)}^{}t-4\int \frac{t^2}{1+t^2}dt\\ & =2t\ln (1+t^2)-4\int \frac{t}{t+t^2}dt\\ & =2t\ln (1+t^2)-4\int 1dt+4\int \frac{1}{1+t^2}dt\\ & =2t\ln (1+t^2)-4t+4\arctan t+C\\ & =2\sqrt{x}\ln (1+x)-4\sqrt{x}+4\arctan \sqrt{x}+C\\ & \end{array}$$

答案：

$$\begin{array}{rl}& \int \frac{{\ln }^{2}x}{x^2}\mathrm{d}x\\ & \\ & =-\int {\ln }^{2}x\mathrm{d}\frac{1}{x}\\ & \\ & =-\frac{1}{x}{\ln }^{2}x+\int \frac{1}{x}2\ln x\frac{1}{x}\mathrm{d}x\\ & \\ & =-\frac{1}{x}{\ln }^{2}x-2\int \ln x\mathrm{d}\frac{1}{x}\\ & \\ & =-\frac{1}{x}{\ln }^{2}x-2\frac{1}{x}\ln x+2\int \frac{1}{x}.\frac{1}{x}dx\\ & \\ & =-\frac{1}{x}{\ln }^{2}x-2\frac{1}{x}\ln x+2\int x^{-2}dx\\ & \\ & =-\frac{1}{x}{\ln }^{2}x-2\frac{1}{x}\ln x-2\frac{1}{x}+C\end{array}$$

答案：

$$\begin{array}{rl}\text{原式}& =\int (\ln x\cdot {\mathrm{e}}^x+\frac{1}{x}\cdot {\mathrm{e}}^x)\mathrm{d}x=\int \ln x\mathrm{d}{\mathrm{e}}^x+\int \frac{1}{x}{\mathrm{e}}^x\mathrm{d}x\\ & =\ln x\cdot {\mathrm{e}}^x-\int {\mathrm{e}}^x\cdot \frac{1}{x}\mathrm{d}x+\int \frac{1}{x}{\mathrm{e}}^x\mathrm{d}x=\ln x\cdot {\mathrm{e}}^x+C.\end{array}$$

答案：

$$\begin{array}{rl}\text{原式}& =\int \frac{{\mathrm{e}}^x(1+2\sin \frac{x}{2}\mathrm{c}\mathrm{o}\mathrm{s}\frac{x}{2})}{2{\cos }^2\frac{x}{2}}\mathrm{d}x=\int {\mathrm{e}}^x\cdot \frac{1}{2}{\sec }^2\frac{x}{2}\mathrm{d}x+\int {\mathrm{e}}^x\cdot \tan \frac{x}{2}\mathrm{d}x\\ & =\int {\mathrm{e}}^x\cdot \frac{1}{2}{\sec }^2\frac{x}{2}\mathrm{d}x+\int \tan \frac{x}{2}\mathrm{d}{\mathrm{e}}^x=\int {\mathrm{e}}^x\cdot \frac{1}{2}{\sec }^2\frac{x}{2}\mathrm{d}x+{\mathrm{e}}^x\tan \frac{x}{2}-\int {\mathrm{e}}^x\cdot \frac{1}{2}{\sec }^2\frac{x}{2}\mathrm{d}x={\mathrm{e}}^x\tan \frac{x}{2}+C.\end{array}$$

答案：

$$\begin{array}{rl}& \int e^{2x}(\tan x+1{)}^{2}dx\\ =& \int e^{2x}({\tan }^{2}x+1+2\tan x)dx\\ =& \int e^{2x}({\tan }^{2}x+1)+2e^{2x}\tan xdx\\ =& \int e^{2x}\cdot {\sec }^{2}xdx+2\int e^{2x}\tan xdx\\ =& \int e^{2x}d\tan x+\int \tan xd{e}^{2x}\\ =& e^{2x}\tan x-\int \tan xd{e}^{2x}+\int \tan xd{e}^{2x}\\ =& e^{2x}\tan x+c\end{array}$$

答案：

$$\begin{array}{rl}& =\int \frac{x^2+1-1}{x^2+1}\arctan x\mathrm{d}x\\ & =\int \arctan x\mathrm{d}x-\int \frac{1}{1+x^2}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}x\mathrm{d}x\\ & =x\arctan x-\int \frac{x}{1+x^2}\mathrm{d}x-\int \arctan x\mathrm{d}(\arctan x)\\ & =x\arctan x-\frac{1}{2}\mathrm{l}\mathrm{n}(1+x^2)-\frac{1}{2}{\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}}^{2}x+C.\end{array}$$

答案：

$$\begin{array}{rl}& =\frac{1}{2}\int \frac{(2x+4)-2}{x^2+4x+13}\mathrm{d}x\\ & =\frac{1}{2}\int \frac{2x+4}{x^2+4x+13}\mathrm{d}x-\int \frac{1}{x^2+4x+13}\mathrm{d}x\\ & =\frac{1}{2}\mathrm{l}\mathrm{n}|x^2+4x+13|-\int \frac{1}{{(x+2)}^2+3^2}\mathrm{d}(x+2)\\ & =\frac{1}{2}\mathrm{l}\mathrm{n}\mid x^2+4x+13\mid -\frac{1}{3}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\frac{x+2}{3}+C.\end{array}$$

答案：

$$\begin{array}{rl}& =\int [\frac{1}{2(x-1)}+\frac{1}{2(x+1)}-\frac{1}{{(x+1)}^2}]\mathrm{d}x\\ & =\frac{1}{2}\mathrm{l}\mathrm{n}\mid x-1\mid +\frac{1}{2}\mathrm{l}\mathrm{n}\mid x+1\mid +\frac{1}{x+1}+C\\ & =\frac{1}{2}\mathrm{l}\mathrm{n}\mid x^2-1\mid +\frac{1}{x+1}+C.\end{array}$$

答案：

$$\begin{array}{l}=\int \frac{x^2(x^3-4x)+x(x^3-4x)+4(x^3-4x)+4x^2+16x-8}{x^3-4x}\mathrm{d}x\\ =\int [x^2+x+4+\frac{4x^2+16x-8}{x(x-2)(x+2)}]\mathrm{d}x\\ =\int (x^2+x+4+\frac{2}{x}+\frac{5}{x-2}-\frac{3}{x+2})\mathrm{d}x\\ =\frac{1}{3}{x}^3+\frac{1}{2}{x}^2+4x+2\ln |x|+5\ln |x-2|-3\ln |x+2|+C.\end{array}$$

答案：

$$\begin{array}{rl}\int {\ln }^2(x+\sqrt{1+x^2})dx& \\ & =x{\ln }^2(x+\sqrt{1+x^2})-\int x\cdot 2\ln (x+\sqrt{1+x^2})\cdot \frac{1}{\sqrt{1+x^2}}dx\\ & =x{\ln }^2(x+\sqrt{1+x^2})-2\int \ln (x+\sqrt{1+x^2})d\sqrt{1+x^2}\\ & =x{\ln }^2(x+\sqrt{1+x^2})-2\sqrt{x^2+1}\ln (x+\sqrt{1+x^2})+{\int }_{}\sqrt{1+x^2}+\frac{1}{\sqrt{1+x^2}}dx\\ & =x{\ln }^2(x+\sqrt{1+x^2})-2\sqrt{x^2+1}\ln (x+\sqrt{1+x^2})+x+\mathbb{C}\end{array}$$

答案：

$$\begin{array}{rl}& =2\int \arcsin \sqrt{x}\mathrm{d}\sqrt{x}+2\int \ln x\mathrm{d}\sqrt{x}\\ & =2\arcsin \sqrt{x}\cdot \sqrt{x}-2\int \frac{\sqrt{x}}{\sqrt{1-x}}\cdot \frac{1}{2\sqrt{x}}\mathrm{d}x+2\ln x\cdot \sqrt{x}-2\int \sqrt{x}\cdot \frac{1}{x}\mathrm{d}x\\ & =2\arcsin \sqrt{x}\cdot \sqrt{x}-\int \frac{1}{\sqrt{1-x}}\mathrm{d}x+2\ln x\cdot \sqrt{x}-2\int \frac{1}{\sqrt{x}}\mathrm{d}x\\ & =2\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}\sqrt{x}\cdot \sqrt{x}+2\sqrt{1-x}+2\mathrm{l}\mathrm{n}x\cdot \sqrt{x}-4\sqrt{x}+C.\end{array}$$

答案：

$$\begin{array}{rl}& \int \frac{dx}{\sin 2x+2{\sin }^{2}x}\\ & =\int \frac{1}{2\sin x\cos x+2{\sin }^{}x}dx\\ & =\int \frac{1}{2\sin (\cos x+1)}dx\\ & =\int \frac{1}{4\sin x\cos x^2}dx\\ & =\int \frac{1}{4\sin x}dtanx\\ & \because \frac{1+\cos x}{2}={\cos }^{2}x\\ & =\frac{1}{4}\int \csc xd\tan x\\ & =\frac{1}{4}\csc x\tan x+\frac{1}{4}\int \tan x\cdot \csc x\cdot \cot xdx\\ & =\frac{1}{4}\csc x\tan x+\frac{1}{4}\int \csc xdx\\ & =\frac{1}{4}\csc x\tan x+\frac{1}{4}\ln |\csc x-\cot x|+c\end{array}$$