高数不定积分72题解答

题目来源：这72道积分题目会积了，绝对是高高手

题目作者： 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

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

第十六题

$$\begin{array}{rl}\text{原式}& =\int \frac{1}{a^2+x^2}dx\\ & =\frac{1}{a}\mathrm{arc}\tan \frac{x}{a}+C\end{array}$$

第十七题

$$\begin{array}{rl}\text{原式}& =\int \frac{1}{a^2-x^2}dx\\ & =\frac{1}{2a}\ln \mid \ln \frac{x-a}{x+a}\mid +C\end{array}$$

第十八题

$$\begin{array}{rl}\text{原式}& =\int \frac{1}{\sqrt{a^2-x^2}}dx\\ & =\mathrm{arc}\sin \frac{x}{a}+C\end{array}$$

第十九题

$$\begin{array}{rl}\text{原式}& =\int {\sin }^{3}xdx\\ & =-\int {\sin }^{2}xd\cos x\\ & =-\int 1-{\cos }^{2}xd\cos x\\ & =-\cos x+\frac{1}{3}{\cos }^{3}x+C\end{array}$$

第二十题

$$\begin{array}{rl}\text{原式}& =\int {\sin }^{5}xdx\\ & =-\int {\sin }^{4}xd\cos x\\ & =-\int {\left(1-{\cos }^{2}x\right)}^{2}d\cos x\\ & =-\int \left(1-2{\cos }^{2}x+{\cos }^{4}x\right)d\cos x\\ & =-\cos x+\frac{2}{3}{\cos }^{3}x+\frac{1}{5}{\cos }^{5}x+C\end{array}$$

第二十一题

$$\begin{array}{rl}\text{原式}& =\int {\cos }^{3}xdx\\ & =\int {\cos }^{2}xd\sin x\\ & =\int \left(1-{\sin }^{2}x\right)d\sin x\\ & =\sin x-\frac{1}{3}{\sin }^{3}x+C\end{array}$$

第二十二题

$$\begin{array}{rl}\text{原式}& =\int {\sin }^{4}xdx\\ & =-\int {\left(\frac{1-\cos 2x}{2}\right)}^{2}dx\\ & =\frac{1}{4}\int {\left(1-\cos 2x\right)}^{2}dx\\ & =\frac{1}{4}\int \left(1-2\cos 2x++{\cos }^{2}2x\right)dx\\ & =\frac{1}{4}x+\int {\cos }^{2}2xdx-\frac{1}{4}\sin 2x\\ & =\frac{1}{4}x-\frac{1}{4}\sin 2x+\frac{1}{4}\int \frac{\cos 4x+1}{2}dx\\ & =\frac{1}{4}x-\frac{1}{4}\sin 2x+\frac{1}{8}\int \left(\cos 4x+1\right)dx\\ & =\frac{1}{4}x-\frac{1}{4}\sin 2x+\frac{1}{8}x+\frac{1}{8}\int \cos 4xdx\\ & =\frac{3}{8}x-\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x+C\end{array}$$

第二十三题

$$\begin{array}{rl}\text{原式}& =\int {\sin }^{2}x{\cos }^{5}xdx\\ & =\int {\sin }^{2}x{\left(1-{\sin }^{2}x\right)}^{2}d\sin x\\ & =\int \left({\sin }^{2}x+{\sin }^{6}x-2{\sin }^{4}x\right)d\sin x\\ & =\frac{1}{3}{\sin }^{3}x+\frac{1}{7}{\sin }^{7}x-\frac{2}{5}{\sin }^{5}x+C\end{array}$$

第二十四题

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

第二十五题

$$\begin{array}{rl}\text{原式}& =\int {\sec }^{3}x{\tan }^{5}xdx\\ & =\int \frac{{\sin }^{5}x}{{\cos }^{8}x}dx\\ & =-\int \frac{{\left(1-{\cos }^{2}x\right)}^2}{{\cos }^{8}x}d\cos x\\ & =-\int \frac{{\left(1-t^2\right)}^2}{t^8}dt\\ & =-\int \left(\frac{1}{t^8}+\frac{1}{t^4}-\frac{2}{t^6}\right)dt\\ & =\frac{1}{7t^7}+\frac{1}{3t^3}-\frac{2}{5t^5}+C\\ & =\frac{1}{7{\cos }^{7}x}+\frac{1}{3{\cos }^{3}x}-\frac{2}{5{\cos }^{5}x}+C\end{array}$$

第二十六题

$$\begin{array}{rl}\text{原式}& =\int {\tan }^{5}x{\sec }^{4}xdx\\ & =\int \frac{{\sin }^{5}x}{{\cos }^{9}x}dx\\ & =-\int \frac{{\sin }^{4}x}{{\cos }^{9}x}d\cos x\\ & =-\int \frac{{\left(1-{\cos }^{2}x\right)}^2}{{\cos }^{9}x}d\cos x\\ & =-\int \frac{1-2{\cos }^{2}x+{\cos }^{4}x}{{\cos }^{9}x}d\cos x\\ & =-\left(\int \frac{1}{{\cos }^{9}x}d\cos x-2\int \frac{1}{{\cos }^{7}x}d\cos x+\int \frac{1}{{\cos }^{5}x}d\cos x\right)\\ & =\frac{1}{8{\cos }^{8}x}-\frac{1}{3{\cos }^{6}x}+\frac{1}{4{\cos }^{4}x}+C\end{array}$$

第二十七题