《高等数学》习题3.2选做
Elegia
2021-06-17 23:30:31
求下列不定积分：
1. $\displaystyle \int x\ln x\mathrm{d} x$
解：$\displaystyle =\frac{1}{2}\int \ln x\mathrm{d}\left( x^{2}\right) =\frac{1}{2}\left[ x^{2}\ln x-\int x^{2}\mathrm{d}(\ln x)\right] =\frac{1}{2}\left[ x^{2}\ln x-\int x\mathrm{d} x\right] =\frac{2\ln x-1}{4} x^{2} +C$
2. $\displaystyle \int x^{2}\mathrm{e}^{ax}\mathrm{d} x$
解：$\displaystyle a=0$ 时为 $\displaystyle \frac{1}{3} x^{3} +C$，否则
$$
\begin{aligned}
& =\frac{1}{a}\int \mathrm{e}^{ax}\mathrm{d}\left( x^{2}\right)\\
& =\frac{1}{a}\mathrm{e}^{ax} x^{2} -\frac{2}{a}\int \mathrm{e}^{ax} x\mathrm{d} x\\
& =\frac{1}{a}\mathrm{e}^{ax} x^{2} -\frac{2}{a^{2}}\int x\mathrm{d}\left(\mathrm{e}^{ax}\right)\\
& =\frac{1}{a}\mathrm{e}^{ax} x^{2} -\frac{2}{a^{2}}\mathrm{e}^{ax} x+\frac{2}{a^{2}}\int \mathrm{e}^{ax}\mathrm{d} x\\
& =\frac{\mathrm{e}^{ax}}{a^{3}}\left( a^{2} x^{2} -2ax+2\right) +C
\end{aligned}
$$
3. $\displaystyle \int x\sin 2x\mathrm{d} x$
解：$\displaystyle =-\frac{1}{2}\int x\mathrm{d}(\cos 2x) =-\frac{1}{2} x\cos 2x+\frac{1}{2}\int \cos 2x\mathrm{d} x=-\frac{1}{2} x\cos 2x+\frac{1}{4}\sin 2x+C$
4. $\displaystyle \int \arcsin x\mathrm{d} x$
解：$\displaystyle =x\arcsin x-\int \frac{x}{\sqrt{1-x^{2}}}\mathrm{d} x=x\arcsin x+\frac{1}{2}\int \frac{\mathrm{d}\left( 1-x^{2}\right)}{\sqrt{1-x^{2}}} =x\arcsin x+\sqrt{1-x^{2}} +C$
5. $\displaystyle \int \arctan x\mathrm{d} x$
解：$\displaystyle =x\arctan x-\int \frac{x}{1+x^{2}}\mathrm{d} x=x\arctan x-\frac{1}{2}\int \frac{\mathrm{d}\left( 1+x^{2}\right)}{1+x^{2}} =x\arctan x-\frac{1}{2}\ln\left| 1+x^{2}\right| +C$
6. $\displaystyle \int \mathrm{e}^{2x}\cos 3x\mathrm{d} x$
解：
$$
\begin{aligned}
I &= \int \mathrm{e}^{2x}\cos 3x\mathrm{d} x\\
& =\frac{1}{2}\int \cos 3x\mathrm{d}\left(\mathrm{e}^{2x}\right)\\
& =\frac{1}{2}\mathrm{e}^{2x}\cos 3x-\frac{1}{2}\int \mathrm{e}^{2x}\mathrm{d}(\cos 3x)\\
& =\frac{1}{2}\mathrm{e}^{2x}\cos 3x+\frac{3}{2}\int \mathrm{e}^{2x}\sin 3x\mathrm{d} x\\
J & =\int \mathrm{e}^{2x}\sin 3x\mathrm{d} x\\
& =\frac{1}{2}\int \sin 3x\mathrm{d}\left( \mathrm{e}^{2x}\right)\\
& =\frac{1}{2}\mathrm{e}^{2x}\sin 3x-\frac{1}{2}\int \mathrm{e}^{2x}\mathrm{d}(\sin 3x)\\
& =\frac{1}{2}\mathrm{e}^{2x}\sin 3x-\frac{3}{2}\int \mathrm{e}^{2x}\cos 3x\mathrm{d} x\\
& =\frac{1}{2}\mathrm{e}^{2x}\sin 3x-\frac{3}{2} I+C\\
I & =\mathrm{e}^{2x}\left(\frac{1}{2}\cos 3x+\frac{3}{4}\sin 3x\right) -\frac{9}{4} I+C\\
I & =\frac{\mathrm{e}^{2x}}{13}( 2\cos 3x+3\sin 3x) +C
\end{aligned}
$$
8. $\displaystyle \int \mathrm{e}^{ax}\sin bx\mathrm{d} x$
解：
$$
\begin{aligned}
I & =\int \mathrm{e}^{ax}\sin bx\mathrm{d} x\\
& =\frac{1}{a}\int \sin bx\mathrm{d}\left(\mathrm{e}^{ax}\right)\\
& =\frac{1}{a}\mathrm{e}^{ax}\sin bx-\frac{b}{a}\int \mathrm{e}^{ax}\cos bx\mathrm{d} x\\
& =\frac{1}{a}\mathrm{e}^{ax}\sin bx-\frac{b}{a} J+C\\
J & =\int \mathrm{e}^{ax}\cos bx\mathrm{d} x\\
& =\frac{1}{a}\int \cos bx\mathrm{d}\left(\mathrm{e}^{ax}\right)\\
& =\frac{1}{a}\mathrm{e}^{ax}\cos bx+\frac{b}{a}\int \mathrm{e}^{ax}\sin bx\mathrm{d} x\\
& =\frac{1}{a}\mathrm{e}^{ax}\cos bx+\frac{b}{a} I+C\\
I & =\frac{1}{a}\mathrm{e}^{ax}\sin bx-\frac{b}{a^{2}}\mathrm{e}^{ax}\cos bx-\frac{b^{2}}{a^{2}} I+C\\
I & =\frac{\mathrm{e}^{ax}}{a^{2} +b^{2}}( a\sin bx-b\cos bx) +C
\end{aligned}
$$
9. $\displaystyle \int \sqrt{1+9x^{2}}\mathrm{d} x$
解：
$$
\begin{aligned}
I & =x\sqrt{1+9x^{2}} -\int \frac{9x^{2}}{\sqrt{1+9x^{2}}}\mathrm{d} x\\
& =x\sqrt{1+9x^{2}} -\int \frac{\left( 1+9x^{2}\right) -1}{\sqrt{1+9x^{2}}}\mathrm{d} x\\
& =x\sqrt{1+9x^{2}} -I+\frac{1}{3}\int \frac{1}{\sqrt{1+( 3x)^{2}}}\mathrm{d}( 3x)\\
2I & =x\sqrt{1+9x^{2}} +\frac{1}{3}\ln\left| 3x+\sqrt{1+9x^{2}}\right| +C\\
I & =\frac{1}{2} x\sqrt{1+9x^{2}} +\frac{1}{6}\ln\left| 3x+\sqrt{1+9x^{2}}\right| +C
\end{aligned}
$$
10. $\displaystyle \int x\cosh x\mathrm{d} x$
解：
$$
\begin{aligned}
& =\int x\mathrm{d}(\sinh x)\\
& =x\sinh x-\int \sinh x\mathrm{d} x\\
& =x\sinh x-\cosh x+C
\end{aligned}
$$
11. $\displaystyle \int \ln\left( x+\sqrt{1+x^{2}}\right)\mathrm{d} x$
解：
$$
\begin{aligned}
& =x\ln\left( x+\sqrt{1+x^{2}}\right) -\int \frac{x}{x+\sqrt{1+x^{2}}} \cdotp \left( 1+\frac{x}{\sqrt{1+x^{2}}}\right)\mathrm{d} x\\
& =x\ln\left( x+\sqrt{1+x^{2}}\right) -\int \frac{x}{\sqrt{1+x^{2}}}\mathrm{d} x\\
& =x\ln\left( x+\sqrt{1+x^{2}}\right) -\frac{1}{2}\int \frac{\mathrm{d}\left( 1+x^{2}\right)}{\sqrt{1+x^{2}}}\\
& =x\ln\left( x+\sqrt{1+x^{2}}\right) -\sqrt{1+x^{2}} +C
\end{aligned}
$$
12. $\displaystyle \int (\arccos x)^{2}\mathrm{d} x$
解：先换元，令 $\displaystyle x=\cos t$
$$
\begin{aligned}
& =\int t^{2}\mathrm{d}(\cos t)\\
& =t^{2}\cos t-2\int t\cos t\mathrm{d} t\\
& =t^{2}\cos t-2\int t\mathrm{d}(\sin t)\\
& =t^{2}\cos t-2t\sin t+2\int \sin t\mathrm{d} t\\
& =t^{2}\cos t-2t\sin t-2\cos t+C\\
& =x\arccos^{2} x-2\sqrt{1-x^{2}}\arccos x-2x+C
\end{aligned}
$$
13. $\displaystyle \int \frac{x\arccos x}{\left( 1-x^{2}\right)^{2}}\mathrm{d} x$
解：
$$
\begin{aligned}
& =-\frac{1}{2}\int \frac{\arccos x}{\left( 1-x^{2}\right)^{2}}\mathrm{d}\left( 1-x^{2}\right)\\
& =\frac{1}{2}\int \arccos x\mathrm{d}\left(\frac{1}{1-x^{2}}\right)\\
& =\frac{\arccos x}{2\left( 1-x^{2}\right)} -\frac{1}{2}\int \frac{\mathrm{d}(\arccos x)}{1-x^{2}}\\
& =\frac{\arccos x}{2\left( 1-x^{2}\right)} +\frac{1}{2}\int \frac{\mathrm{d} x}{\left( 1-x^{2}\right)^{3/2}}
\end{aligned}
$$
令 $\displaystyle x=\sin t$，有
$$
\begin{aligned}
\int \frac{\mathrm{d} x}{\left( 1-x^{2}\right)^{3/2}} & =\int \frac{\cos t}{\cos^{3} t}\mathrm{d} t\\
& =\int \sec^{2} t\\
& =\tan t+C\\
& =\frac{x}{\sqrt{1-x^{2}}} +C
\end{aligned}
$$
因此原式 $\displaystyle =\frac{\arccos x}{2\left( 1-x^{2}\right)} +\frac{x}{2\sqrt{1-x^{2}}} +C$
14. $\displaystyle \int \arctan\sqrt{x}\mathrm{d} x$
解：
$$
\begin{aligned}
& =x\arctan\sqrt{x} -\int \frac{x}{1+x}\mathrm{d}\left(\sqrt{x}\right)\\
& =x\arctan\sqrt{x} -\int \left( 1-\frac{1}{1+x}\right)\mathrm{d}\left(\sqrt{x}\right)\\
& =( x+1)\arctan\sqrt{x} -\sqrt{x} +C
\end{aligned}
$$
15. $\displaystyle \int \frac{\arcsin x}{x^{2}}\mathrm{d} x$
解：
$$
\begin{aligned}
& =-\int \arcsin x\mathrm{d}\left( x^{-1}\right)\\
& =-\frac{\arcsin x}{x} +\int \frac{\mathrm{d}(\arcsin x)}{x}\\
& =-\frac{\arcsin x}{x} +\int \frac{\mathrm{d} x}{x\sqrt{1-x^{2}}}
\end{aligned}
$$
令 $\displaystyle x=\sin t$，有
$$
\begin{aligned}
\int \frac{\mathrm{d} x}{x\sqrt{1-x^{2}}} & =\int \frac{\cos t\mathrm{d} t}{\sin t\cos t}\\
& =\int \frac{\mathrm{d} t}{\sin t}\\
& =-\int \frac{\mathrm{d}(\cos t)}{1-\cos^{2} t}\\
& =-\frac{1}{2}\log\left| \frac{1+\cos t}{1-\cos t}\right| +C\\
& =-\frac{1}{2}\log\frac{1+\sqrt{1-x^{2}}}{1-\sqrt{1-x^{2}}} +C\\
& =-\frac{1}{2}\log\frac{\left( 1+\sqrt{1-x^{2}}\right)^{2}}{x^{2}} +C\\
& =-\log\left| 1+\sqrt{1-x^{2}}\right| +\log| x| +C
\end{aligned}
$$
因此结果为 $\displaystyle -\frac{\arcsin x}{x} -\log\left| 1+\sqrt{1-x^{2}}\right| +\log| x| +C$
16. $\displaystyle \int x^{3}(\ln x)^{2}\mathrm{d} x$
解：
$$
\begin{aligned}
I_{n} & =\int x^{3}(\ln x)^{n}\mathrm{d} x\\
& =\frac{1}{4}\int \ln^{n} x\mathrm{d}\left( x^{4}\right)\\
& =\frac{1}{4} x^{4}\ln^{n} x-\frac{1}{4}\int x^{4} \cdotp \frac{n\ln^{n-1} x}{x}\mathrm{d} x\\
& =\frac{1}{4} x^{4}\ln^{n} x-\frac{n}{4} I_{n-1}\\
I_{2} & =\frac{1}{4} x^{4}\ln^{2} x-\frac{1}{2} I_{1}\\
& =\frac{1}{4} x^{4}\ln^{2} x-\frac{1}{8} x^{4}\ln x+\frac{1}{8} I_{0}\\
& =\frac{1}{4} x^{4}\ln^{2} x-\frac{1}{8} x^{4}\ln x+\frac{1}{32} x^{4} +C\\
& =\frac{1}{4} x^{4}\left(\ln^{2} x-\frac{1}{2}\ln x+\frac{1}{8}\right) +C
\end{aligned}
$$
17. $\displaystyle \int \frac{x\arctan x}{\left( 1+x^{2}\right)^{5/2}}\mathrm{d} x$
解：令 $\displaystyle x=\tan t$，有
$$
\begin{aligned}
& =\int \frac{t\tan t}{\left( 1+\tan^{2} t\right)^{5/2}}\sec^{2} t\mathrm{d} t\\
& =\int \frac{t\tan t}{\sec^{5} t}\sec^{2} t\mathrm{d} t\\
& =\int t\tan t\cdot \cos^{3} t\mathrm{d} t\\
& =-\int t\cos^{2} t\mathrm{d}(\cos t)\\
& =-\frac{1}{3}\int t\mathrm{d}\left(\cos^{3} t\right)\\
& =-\frac{1}{3} t\cos^{3} t+\frac{1}{3}\int \cos^{3} t\mathrm{d} t\\
& =-\frac{1}{3} t\cos^{3} t+\frac{1}{3}\int \left( 1-\sin^{2} t\right)\mathrm{d}(\sin t)\\
& =-\frac{1}{3} t\cos^{3} t-\frac{1}{9}\sin^{3} t+\frac{1}{3}\sin t+C\\
& =-\frac{1}{3}\left( 1+x^{2}\right)^{-3/2}\arctan x-\frac{x^{3}}{9}\left( 1+x^{2}\right)^{-3/2} +\frac{x}{3\sqrt{1+x^{2}}} +C
\end{aligned}
$$
18. $\displaystyle \int x\ln\left( x+\sqrt{1+x^{2}}\right)\mathrm{d} x$
解：根据第 11 题我们有
$$
\begin{aligned}
I & =\int x\mathrm{d}\left[ x\ln\left( x+\sqrt{1+x^{2}}\right) -\sqrt{1+x^{2}}\right]\\
& =x\left[ x\ln\left( x+\sqrt{1+x^{2}}\right) -\sqrt{1+x^{2}}\right] -\int \left[ x\ln\left( x+\sqrt{1+x^{2}}\right) -\sqrt{1+x^{2}}\right]\mathrm{d} x\\
& =x\left[ x\ln\left( x+\sqrt{1+x^{2}}\right) -\sqrt{1+x^{2}}\right] -I+\int \sqrt{1+x^{2}}\mathrm{d} x\\
J & =\int \sqrt{1+x^{2}}\mathrm{d} x\\
& =x\sqrt{1+x^{2}} -\int \frac{\left( 1+x^{2}\right) -1}{\sqrt{1+x^{2}}}\mathrm{d} x\\
& =x\sqrt{1+x^{2}} -J+\int \frac{\mathrm{d} x}{\sqrt{1+x^{2}}}\\
2J & =x\sqrt{1+x^{2}} +\ln\left( x+\sqrt{1+x^{2}}\right) +C\\
J & =\frac{1}{2} x\sqrt{1+x^{2}} +\frac{1}{2}\ln\left( x+\sqrt{1+x^{2}}\right) +C\\
I & =\frac{1}{2} x\left[ x\ln\left( x+\sqrt{1+x^{2}}\right) -\sqrt{1+x^{2}}\right] +\frac{1}{4}\left[ x\sqrt{1+x^{2}} +\ln\left( x+\sqrt{1+x^{2}}\right)\right] +C\\
& =\frac{2x^{2} +1}{4}\ln\left( x+\sqrt{1+x^{2}}\right) -\frac{1}{4} x\sqrt{1+x^{2}} +C
\end{aligned}
$$