《高等数学》习题3.1
Elegia
2021-06-17 15:30:20
求下列不定积分：
1. $\displaystyle \int \sqrt{1+2x}\mathrm{d} x$
解：$\displaystyle =\int \sqrt{1+2x} \cdotp \frac{1}{2}\mathrm{d}( 1+2x) =\frac{1}{3}( 1+2x)^{3/2} +C$
2. $\displaystyle \int \frac{3x}{\left( x^{2} +1\right)^{2}}\mathrm{d} x$
解：$\displaystyle =\int \frac{3/2}{\left( x^{2} +1\right)^{2}}\mathrm{d}\left( x^{2} +1\right) =\frac{-3}{2\left( x^{2} +1\right)} +C$
3. $\displaystyle \int x\sqrt{2x^{2} +7}\mathrm{d} x$
解：$\displaystyle =\int \sqrt{2x^{2} +7} \cdotp \frac{1}{4}\mathrm{d}\left( 2x^{2} +7\right) =\frac{1}{6}\left( 2x^{2} +7\right)^{3/2} +C$
4. $\displaystyle \int \left( 2x^{3/2} +1\right)^{2/3}\sqrt{x}\mathrm{d} x$
解：$\displaystyle =\int \left( 2x^{3/2} +1\right)^{2/3} \cdot \frac{1}{3}\mathrm{d}\left( 2x^{3/2} +1\right) =\frac{1}{5}\left( 2x^{3/2} +1\right)^{5/3} +C$
5. $\displaystyle \int \frac{\mathrm{e}^{1/x}}{x^{2}}\mathrm{d} x$
解：$\displaystyle =\int \mathrm{e}^{1/x} \cdotp -\mathrm{d}\left(\frac{1}{x}\right) =-\mathrm{e}^{1/x} +C$
6. $\displaystyle \int \frac{\mathrm{d} x}{( 2-x)^{100}}$
解：$\displaystyle =\int \frac{-\mathrm{d}( 2-x)}{( 2-x)^{100}} =\frac{1}{99( 2-x)^{99}} +C$
7. $\displaystyle \int \frac{\mathrm{d} x}{3+5x^{2}}$
解：$\displaystyle =\frac{1}{3}\int \frac{\mathrm{d} x}{1+\frac{5}{3} x^{2}} =\frac{1}{3}\int \frac{\frac{\sqrt{15}}{5}\mathrm{d}\left(\frac{\sqrt{15}}{3} x\right)}{1+\left(\frac{\sqrt{15}}{3} x\right)^{2}} =\frac{\sqrt{15}}{15}\arctan\frac{\sqrt{15}}{3} x+C$
8. $\displaystyle \int \frac{\mathrm{d} x}{\sqrt{7-3x^{2}}}$
解：$\displaystyle =\frac{1}{\sqrt{7}}\int \frac{\mathrm{d} x}{\sqrt{1-\left(\frac{\sqrt{3}}{\sqrt{7}} x\right)^{2}}} =\frac{1}{\sqrt{3}}\int \frac{\mathrm{d}\left(\frac{\sqrt{3}}{\sqrt{7}} x\right)}{\sqrt{1-\left(\frac{\sqrt{3}}{\sqrt{7}} x\right)^{2}}} =\frac{1}{\sqrt{3}}\arcsin\frac{\sqrt{21}}{7} x+C$
9. $\displaystyle \int \frac{\mathrm{d} x}{\sqrt{x}( 1+x)}$
解：$\displaystyle =\int \frac{2\mathrm{d}\left(\sqrt{x}\right)}{\left( 1+\sqrt{x}^{2}\right)} =2\arctan\sqrt{x} +C$
10. $\displaystyle \int \frac{\mathrm{e}^{x}}{2+\mathrm{e}^{2x}}\mathrm{d} x$
解：$\displaystyle =\frac{1}{\sqrt{2}}\int \frac{\mathrm{d}\left(\frac{1}{\sqrt{2}}\mathrm{e}^{x}\right)}{1+\left(\frac{1}{\sqrt{2}}\mathrm{e}^{x}\right)^{2}} =\frac{1}{\sqrt{2}}\arctan\frac{\mathrm{e}^{x}}{\sqrt{2}} +C$
11. $\displaystyle \int \frac{\mathrm{d} x}{\sqrt{\mathrm{e}^{-2x} -1}}$
解：$\displaystyle =\int \frac{\mathrm{e}^{x}\mathrm{d} x}{\sqrt{1-\mathrm{e}^{2x}}} =\int \frac{\mathrm{d}\left(\mathrm{e}^{x}\right)}{\sqrt{1-\left(\mathrm{e}^{x}\right)^{2}}} =\arcsin\mathrm{e}^{x} +C$
12. $\displaystyle \int \frac{\mathrm{d} x}{\mathrm{e}^{x} -\mathrm{e}^{-x}}$
解：$\displaystyle =\int \frac{\mathrm{d}\left(\mathrm{e}^{x}\right)}{\mathrm{e}^{2x} -1} =\frac{1}{2}\int \left(\frac{1}{\mathrm{e}^{x} -1} -\frac{1}{\mathrm{e}^{x} +1}\right)\mathrm{d}\left(\mathrm{e}^{x}\right) =\frac{1}{2}\ln\left| \frac{\mathrm{e}^{x} -1}{\mathrm{e}^{x} +1}\right| +C$
13. $\displaystyle \int \frac{\ln\ln x}{x\ln x}\mathrm{d} x$
解：$\displaystyle =\int \frac{\ln\ln x}{\ln x}\mathrm{d}(\ln x) =\int \ln\ln x\mathrm{d}(\ln\ln x) =\frac{1}{2}(\ln\ln x)^{2} +C$
14. $\displaystyle \int \frac{\mathrm{d} x}{1+\cos x}$
解：$\displaystyle =\int \frac{\mathrm{d} x}{2\cos^{2}\frac{x}{2}} =\int \sec^{2}\frac{x}{2}\mathrm{d}\left(\frac{x}{2}\right) =\tan\frac{x}{2} +C$
15. $\displaystyle \int \frac{\mathrm{d} x}{1-\sin x}$
解：$\displaystyle =\int \frac{\mathrm{d}\left( x+\frac{\pi }{2}\right)}{1+\cos\left( x+\frac{\pi }{2}\right)} =\tan\left(\frac{x}{2} +\frac{\pi }{4}\right) +C$
16. $\displaystyle \int \frac{x^{14}}{\left( x^{5} +1\right)^{4}}\mathrm{d} x$
解：$\displaystyle =\frac{1}{5}\int \frac{x^{10}}{\left( x^{5} +1\right)^{4}}\mathrm{d}\left( x^{5}\right) =\frac{1}{5}\int \frac{\left( x^{5} +1-1\right)^{2}}{\left( x^{5} +1\right)^{4}}\mathrm{d}\left( x^{5} +1\right)$，令 $\displaystyle y=x^{5} +1$ 则有 $\displaystyle =\frac{1}{5}\int \frac{( y-1)^{2}}{y^{4}}\mathrm{d} y=\frac{1}{5}\int \left( y^{-2} -2y^{-3} +y^{-4}\right)\mathrm{d} y=\frac{1}{5}\left( -\frac{1}{y} +\frac{1}{y^{2}} -\frac{1}{3y^{3}}\right) +C=-\frac{1+3x^{5} +3x^{10}}{15\left( 1+x^{5}\right)^{3}} +C$
17. $\displaystyle \int \frac{x^{2n-1}}{x^{n} -1}\mathrm{d} x$
解：$\displaystyle =\frac{1}{n}\int \left( 1+\frac{1}{x^{n} -1}\right)\mathrm{d}\left( x^{n} -1\right) =\frac{1}{n}\left( x^{n} -1+\ln\left| x^{n} -1\right| \right) +C=\frac{1}{n}\left( x^{n} +\ln\left| x^{n} -1\right| \right)$。
18. $\displaystyle \int \frac{\mathrm{d} x}{x\left( x^{5} +2\right)}$
解：$\displaystyle =\int \frac{\mathrm{d}(\ln |x|)}{x^{5} +2}$，换元 $\displaystyle y=x^{5}$，有 $\displaystyle =\frac{1}{5}\int \frac{\mathrm{d}(\ln |y|)}{y+2} =\frac{1}{5}\int \frac{\mathrm{d} y}{y( y+2)} =\frac{1}{10}\int \left(\frac{1}{y} -\frac{1}{y+2}\right)\mathrm{d} y=\frac{1}{10}\ln\left| \frac{y}{y+2}\right| +C=\frac{1}{10}\ln\left| \frac{x^{5}}{x^{5} +2}\right| +C$
19. $\displaystyle \int \frac{\ln( x+2) -\ln x}{x( x+2)}\mathrm{d} x$
解：
$$
\begin{aligned}
& =\frac{1}{2}\int \ln\frac{x+2}{x}\left(\frac{1}{x} -\frac{1}{x+2}\right)\mathrm{d} x\\
& =-\frac{1}{2}\int \ln\frac{x+2}{x}\mathrm{d}\left(\ln\left| \frac{x+2}{x}\right| \right)\\
& =-\frac{1}{2}\int \ln\frac{x+2}{x}\mathrm{d}\left(\ln\frac{x+2}{x}\right)\\
& =-\frac{1}{4}\ln^{2}\frac{x+2}{x} +C
\end{aligned}
$$
20. $\displaystyle \int \frac{\mathrm{e}^{\arctan x} +x\ln\left( 1+x^{2}\right)}{1+x^{2}}\mathrm{d} x$
解：分别计算，第一部分 $\displaystyle \int \frac{\mathrm{e}^{\arctan x}}{1+x^{2}}\mathrm{d} x=\int \mathrm{e}^{\arctan x}\mathrm{d}(\arctan x) =\mathrm{e}^{\arctan x} +C$，第二部分 $\displaystyle \int \frac{x\ln\left( 1+x^{2}\right)}{1+x^{2}}\mathrm{d} x=\frac{1}{2}\int \frac{\ln\left( 1+x^{2}\right)}{1+x^{2}}\mathrm{d}\left( 1+x^{2}\right) =\frac{1}{2}\int \ln\left( 1+x^{2}\right)\mathrm{d}\left[\ln\left( 1+x^{2}\right)\right] =\frac{1}{4}\ln^{2}\left( 1+x^{2}\right) +C$，综上，积分为 $\displaystyle \mathrm{e}^{\arctan x} +\frac{1}{4}\ln^{2}\left( 1+x^{2}\right) +C$。
21. $\displaystyle \int \sin 2x\cos 2x\mathrm{d} x$
解：$\displaystyle =\int \frac{1}{2}\sin 4x\mathrm{d} x=\frac{1}{8}\int \sin 4x\mathrm{d}( 4x) =-\frac{1}{8}\cos 4x+C$
22. $\displaystyle \int \sin^{2}\frac{x}{2}\cos\frac{x}{2}\mathrm{d} x$
解：$\displaystyle =2\int \sin^{2}\frac{x}{2}\mathrm{d}\left(\sin\frac{x}{2}\right) =\frac{2}{3}\sin^{3}\frac{x}{2} +C$
23. $\displaystyle \int \sin 5x\sin 6x\mathrm{d} x$
解：$\displaystyle =\int \frac{1}{2}(\cos x-\cos 11x)\mathrm{d} x=\frac{1}{2}\sin x-\frac{1}{22}\sin 7x+C$
24. $\displaystyle \int \frac{2x-1}{\sqrt{1-x^{2}}}\mathrm{d} x$
解：$\displaystyle \int \frac{2x}{\sqrt{1-x^{2}}}\mathrm{d} x=-\int \frac{\mathrm{d}\left( 1-x^{2}\right)}{\sqrt{1-x^{2}}} =-2\sqrt{1-x^{2}} +C$，$\displaystyle \int \frac{-\mathrm{d} x}{\sqrt{1-x^{2}}} =\arccos x+C$，因此积分 $\displaystyle =\arccos x-2\sqrt{1-x^{2}} +C$。
25. $\displaystyle \int \frac{x^{3} +x}{\sqrt{1-x^{2}}}\mathrm{d} x$
解：$\displaystyle =\frac{1}{2}\int \frac{1+x^{2}}{\sqrt{1-x^{2}}}\mathrm{d}\left( x^{2}\right) =-\frac{1}{2}\int \frac{x^{2} -1+2}{\sqrt{1-x^{2}}}\mathrm{d}\left( 1-x^{2}\right) =\frac{1}{3}\left( 1-x^{2}\right)^{3/2} -2\sqrt{1-x^{2}} +C$
26. $\displaystyle \int \frac{\mathrm{d} x}{\left( a^{2} -x^{2}\right)^{3/2}} ,\quad ( a >0)$
解：令 $\displaystyle x=a\sin t$，有
$$
\begin{aligned}
& =\int \frac{a\cos t\mathrm{d} t}{a^{3}\cos^{3} t}\\
& =\frac{1}{a^{2}}\int \sec^{2} t\mathrm{d} t=\frac{1}{a^{2}}\tan t+C\\
& =\frac{1}{a^{2}}\tan\left(\arcsin\frac{x}{a}\right) +C\\
& =\frac{1}{a^{2}} \cdotp \frac{x/a}{\sqrt{1-( x/a)^{2}}} +C\\
& =\frac{x}{a^{2}\sqrt{a^{2} -x^{2}}} +C
\end{aligned}
$$
27. $\displaystyle \int \frac{\sqrt{x^{2} -a^{2}}}{x}\mathrm{d} x\quad ( a >0)$
解：对于 $\displaystyle x\geq a$，设 $\displaystyle x=a\sec t$，有
$$
\begin{aligned}
& =\int \frac{a\sqrt{\sec^{2} t-1}}{a\sec t} \cdotp \frac{\sin t}{\cos^{2} t}\mathrm{d} t\\
& =\int \sqrt{1-\cos^{2} t}\mathrm{\cdotp \frac{\sin t}{\cos^{2} t} d} t\\
& =\int \tan^{2} t\mathrm{d} t\\
& =\int \left(\sec^{2} t-1\right)\mathrm{d} t\\
& =\tan t-t+C\\
& =\tan\left(\operatorname{arcsec}\frac{x}{a}\right) -\operatorname{arcsec}\frac{x}{a} +C\\
& =\frac{\sqrt{x^{2} -a^{2}}}{a} -\operatorname{arcsec}\frac{x}{a} +C
\end{aligned}
$$
当 $\displaystyle x\leq -a$ 时，令 $\displaystyle x=-a\sec t$，有
$$
\begin{aligned}
& =\tan\left(\operatorname{arcsec}\frac{-x}{a}\right) -\operatorname{arcsec}\frac{-x}{a} +C\\
& =\frac{\sqrt{x^{2} -a^{2}}}{a} -\operatorname{arcsec}\frac{-x}{a} +C
\end{aligned}
$$
综上，可以表为 $\displaystyle \frac{\sqrt{x^{2} -a^{2}}}{a} -\operatorname{arcsec}\left| \frac{x}{a}\right| +C$。
28. $\displaystyle \int \frac{x^{2}}{\sqrt{a^{2} -x^{2}}}\mathrm{d} x\quad ( a >0)$
解：设 $\displaystyle x=a\sin t$，有
$$
\begin{aligned}
& =\int \frac{a^{2}\sin^{2} t}{a\cos t} a\cos t\mathrm{d} t\\
& =\int a^{2}\sin^{2} t\mathrm{d} t\\
& =a^{2}\int \frac{1-\cos 2t}{2}\mathrm{d} t\\
& =\frac{a^{2}}{2} t-\frac{a^{2}}{4}\sin 2t\\
& =\frac{a^{2}}{2}\arcsin\frac{x}{a} -\frac{a^{2}}{4}\sin\left( 2\arcsin\frac{x}{a}\right)\\
& =\frac{a^{2}}{2}\arcsin\frac{x}{a} -\frac{a^{2}}{2} \cdotp \frac{x}{a} \cdotp \sqrt{1-\left(\frac{x}{a}\right)^{2}}\\
& =\frac{a^{2}}{2}\arcsin\frac{x}{a} -\frac{x}{2}\sqrt{a^{2} -x^{2}}
\end{aligned}
$$
29. $\displaystyle \int \frac{\mathrm{d} x}{\sqrt{1+\mathrm{e}^{3x}}}$
解：令 $\displaystyle y=\sqrt{1+\mathrm{e}^{3x}}$，有 $\displaystyle x=\frac{\ln\left( y^{2} -1\right)}{3}$
$$
\begin{aligned}
& =\int \frac{\mathrm{d}\left(\frac{\ln\left( y^{2} -1\right)}{3}\right)}{y}\\
& =\frac{2}{3}\int \frac{y}{y\left( y^{2} -1\right)}\mathrm{d} y\\
& =\frac{1}{3}\int \left(\frac{1}{y-1} -\frac{1}{y+1}\right)\mathrm{d} y\\
& =\frac{1}{3}\ln\left| \frac{y-1}{y+1}\right| +C\\
& =\frac{1}{3}\ln\frac{y-1}{y+1} +C\\
& =\frac{1}{3}\ln\frac{\sqrt{1+\mathrm{e}^{3x}} -1}{\sqrt{1+\mathrm{e}^{3x}} +1} +C\\
& =\frac{1}{3}\ln\frac{\mathrm{e}^{3x}}{\left(\sqrt{1+\mathrm{e}^{3x}} +1\right)^{2}} +C\\
& =x-\frac{2}{3}\ln\left(\sqrt{1+\mathrm{e}^{3x}} +1\right) +C
\end{aligned}
$$
30. $\displaystyle \int \frac{x^{3}}{\sqrt{1+x^{8}}}\mathrm{d} x$
解：
$$
\begin{aligned}
& =\frac{1}{4}\int \frac{\mathrm{d}\left( x^{4}\right)}{\sqrt{1+x^{8}}}\\
& =\frac{1}{4}\ln\left( x^{4} +\sqrt{x^{8} +1}\right) +C
\end{aligned}
$$
31. $\displaystyle \int \frac{\mathrm{d} x}{x^{6}\sqrt{1+x^{2}}}$
解：令 $\displaystyle x=\tan t$，有
$$
\begin{aligned}
& =\int \frac{\sec^{2} t\mathrm{d} t}{\tan^{6} t\sqrt{1+\tan^{2} t}}\\
& =\int \frac{\sec t\mathrm{d} t}{\tan^{6} t}\\
& =\int \frac{\cos^{5} t\mathrm{d} t}{\sin^{6} t}\\
& =\int \frac{\left( 1-\sin^{2} t\right)^{2}\mathrm{d}(\sin t)}{\sin^{6} t}
\end{aligned}
$$
令 $\displaystyle y=\sin t$，为
$$
\begin{aligned}
& =\int \frac{\left( 1-y^{2}\right)^{2}\mathrm{d} y}{y^{6}}\\
& =-\frac{1}{5y^{5}} +\frac{2}{3y^{3}} -\frac{1}{y} +C
\end{aligned}
$$
由于 $\displaystyle y=\sin\arctan x=\frac{x}{\sqrt{1+x^{2}}}$，积分结果为
$$
\begin{aligned}
& \quad -\frac{\sqrt{1+x^{2}}}{x}\left[\frac{1}{5} \cdotp \left(\frac{1+x^{2}}{x^{2}}\right)^{2} -\frac{2}{3} \cdotp \frac{1+x^{2}}{x^{2}} +1\right] +C\\
& =-\frac{\sqrt{1+x^{2}}}{x} \cdot \frac{3-4x^{2} +8x^{4}}{15x^{4}} +C\\
& =-\sqrt{1+x^{2}} \cdot \frac{3-4x^{2} +8x^{4}}{15x^{5}} +C
\end{aligned}
$$
32. $\displaystyle \int \frac{\mathrm{e}^{2x}}{\sqrt[3]{1+\mathrm{e}^{x}}}\mathrm{d} x$
解：$\displaystyle =\int \frac{\mathrm{e}^{x}}{\sqrt[3]{1+\mathrm{e}^{x}}}\mathrm{d}\left(\mathrm{e}^{x}\right) =\int \frac{\left( 1+\mathrm{e}^{x}\right) -1}{\sqrt[3]{1+\mathrm{e}^{x}}}\mathrm{d}\left( 1+\mathrm{e}^{x}\right) =\frac{3}{5}\left( 1+\mathrm{e}^{x}\right)^{5/3} -\frac{3}{2}\left( 1+\mathrm{e}^{x}\right)^{2/3} +C$。
33. $\displaystyle \int \frac{\mathrm{d} x}{\sqrt{3+x-x^{2}}}$
解：$\displaystyle =\int \frac{\mathrm{d} x}{\sqrt{\frac{13}{4} -\left( x-\frac{1}{2}\right)^{2}}} =\int \frac{\mathrm{d}\left(\frac{2}{\sqrt{13}} x-\frac{1}{\sqrt{13}}\right)}{\sqrt{1-\left(\frac{2}{\sqrt{13}} x-\frac{1}{\sqrt{13}}\right)^{2}}} =\arcsin\left(\frac{2}{\sqrt{13}} x-\frac{1}{\sqrt{13}}\right) +C$
34. $\displaystyle \int \sqrt{7+x-x^{2}}\mathrm{d} x$
解：$\displaystyle =\int \sqrt{\frac{29}{4} -\left( x-\frac{1}{2}\right)^{2}}\mathrm{d} x=\frac{29}{8}\arcsin\frac{2x-1}{\sqrt{29}} +\frac{2x-1}{4}\sqrt{7+x-x^{2}} +C$
35. $\displaystyle \int \frac{\mathrm{d} x}{1+\sqrt{x-1}}$
解：设 $\displaystyle x=t^{2} +1$，有
$$
\begin{aligned}
& =\int \frac{2t\mathrm{d} t}{1+t}\\
& =2\int \left( 1-\frac{1}{1+t}\right)\mathrm{d} t\\
& =2t-2\ln| 1+t| +C\\
& =2\sqrt{x-1} -2\ln\left| 1+\sqrt{x-1}\right| +C\\
& =2\sqrt{x-1} -2\ln\left( 1+\sqrt{x-1}\right) +C
\end{aligned}
$$