题目来源：这72道积分题目会积了，绝对是高高手
作者： 湖心亭看雪

第一题

$$\begin{array}{rl}\text{原式}& =\int \frac{1}{5x+3}dx\\ & =\frac{1}{5}\int \frac{1}{5x+3}d(5x+3)\\ & =\frac{1}{5}ln(5x+3)+C\end{array}$$

第二题

$$\begin{array}{rl}\text{原式}& =\int e^{2x+3}dx\\ & =\frac{1}{2}\int e^{2x+3}d(2x+3)\\ & =\frac{1}{2}{e}^{2x+3}+C\end{array}$$

第三题

$$\begin{array}{rl}\text{原式}& =\int x{e}^{x^2}dx\\ & =\frac{1}{2}\int e^{x^2}d{x}^2\\ & =\frac{1}{2}{e}^{x^2}+C\end{array}$$

第四题

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

第五题

$$\begin{array}{rl}\text{原式}& =\int \frac{1}{x^2}sin\frac{1}{x}dx\\ & =-\int sin\frac{1}{x}d(\frac{1}{x})\\ & =cos(\frac{1}{x})+C\end{array}$$

第六题

$$\begin{array}{rl}\text{原式}& =\int \frac{e^{3\sqrt[]{x}}}{\sqrt[]{x}}dx\\ & =2\int \frac{e^{3\sqrt[]{x}}}{2\sqrt{x}}dx\\ & =\frac{2}{3}\int e^{3\sqrt[]{x}}d(3\sqrt[]{x})\\ & =\frac{2}{3}{e}^{3\sqrt[]{x}}+C\end{array}$$

第七题

$$\begin{array}{rl}\text{原式}& =\int \frac{1}{x(1+x^6)}dx\\ & =\int (\frac{1}{x}-\frac{x^5}{1+x^6})dx\\ & =lnx-\frac{1}{6}\int \frac{1}{1+x^6}d(1+x^6)\\ & =lnx-\frac{1}{6}ln(1+x^6)+C\end{array}$$

第八题

$$\begin{array}{rl}\text{原式}& =\int cos2xdx\\ & =\frac{1}{2}\int cos2xd2x\\ & =\frac{1}{2}sin2x+C\end{array}$$

第九题

$$\begin{array}{rl}\text{原式}& =\int \frac{sinx}{\sqrt[]{5+cosx}}dx\\ & =-\int \frac{1}{\sqrt[]{5+cosx}}d(5+cosx)\\ & =-2\int \frac{1}{2\sqrt[]{5+cosx}}d(5+cosx)\\ & =-2\sqrt[]{5+cosx}+C\end{array}$$

第十题

$$\begin{array}{rl}\text{原式}& =\int ta{n}^{4}xdx\\ & =\int (se{c}^{2}x-1{)}^{2}dx\\ & =\int (se{c}^{4}x-2se{c}^{2}x+1)dx\\ & =\int se{c}^{2}x(se{c}^{2}x-2)dx+\int 1dx\\ & =\int se{c}^{2}x(ta{n}^{2}x-1)dx+x\\ & =\int se{c}^{2}xta{n}^{2}xdx-\int se{c}^{2}dx+x\\ & =\int ta{n}^{2}xdtanx-tanx+x\\ & =\frac{1}{3}ta{n}^{3}x-tanx+x+C\end{array}$$

第十一题

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

第十二题

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

第十三题

$$\begin{array}{rl}\text{原式}& =\int \frac{1}{xl{n}^{2}x}\\ & =\int \frac{1}{l{n}^{2}x}dlnx\\ & =-\frac{1}{lnx}+C\end{array}$$

第十四题

$$\begin{array}{rl}\text{原式}& =\int \frac{1}{x(1+2lnx)}dx\\ & =\frac{1}{2}\int \frac{1}{1+2lnx}d(2lnx)\\ & =\frac{1}{2}ln(1+2lnx)+C\end{array}$$

第十五题

$$\begin{array}{rl}\text{原式}& =\int \frac{1}{a^{2}co{s}^{2}x+b^{2}si{n}^{2}x}dx\\ & =\int \frac{\frac{1}{co{s}^{2}x}}{a^2+b^{2}ta{n}^{2}x}dx\\ & =\int \frac{se{c}^{2}x}{a^2+b^{2}ta{n}^{2}x}dx\\ & =\int \frac{1}{a^2+b^{2}ta{n}^{2}x}dtanx\\ & =\frac{1}{ab}\int \frac{1}{1+\frac{b^{2}ta{n}^{2}x}{a^2}}d(\frac{btanx}{a})\\ & =\frac{1}{ab}arctan(\frac{btanx}{a})+C\end{array}$$