《高等数学》习题3.3选做
Elegia
2021-06-19 18:18:06
求下列不定积分：
1. $\displaystyle \int \frac{x-1}{x^{2} +6x+8}\mathrm{d} x$
解：
$$
\begin{aligned}
& =\int \frac{x-1}{( x+2)( x+4)}\mathrm{d} x\\
& =\frac{1}{2}\int \left(\frac{-3}{x+2} +\frac{5}{x+4}\right)\mathrm{d} x\\
& =-\frac{3}{2}\ln| x+2| +\frac{5}{2}\ln| x+4| +C
\end{aligned}
$$
2. $\displaystyle \int \frac{3x^{4} +x^{2} +1}{x^{2} +x-6}\mathrm{d} x$
解：
$$
\begin{aligned}
& =\int \left( 3x^{2} -3x+22+\frac{-40x+133}{x^{2} +x-6}\right)\mathrm{d} x\\
& =x^{3} -\frac{3}{2} x^{2} +22x+\int \frac{-40x+133}{( x+3)( x-2)}\mathrm{d} x\\
& =x^{3} -\frac{3}{2} x^{2} +22x+\frac{1}{5}\int \left(\frac{-253}{x+3} +\frac{53}{x-2}\right)\mathrm{d} x\\
& =x^{3} -\frac{3}{2} x^{2} +22x+\frac{1}{5}( -253\ln| x+3| +53\ln| x-2| ) +C
\end{aligned}
$$
3. $\displaystyle \int \frac{2x^{2} -5}{x^{4} -5x^{2} +6}\mathrm{d} x$
解：
$$
\begin{aligned}
& =\int \frac{2x^{2} -5}{\left( x^{2} -2\right)\left( x^{2} -3\right)}\mathrm{d} x\\
& =\int \left(\frac{1}{x^{2} -2} +\frac{1}{x^{2} -3}\right)\mathrm{d} x\\
& =\int \left[\frac{1}{2\sqrt{2}}\left(\frac{1}{x-\sqrt{2}} -\frac{1}{x+\sqrt{2}}\right) +\frac{1}{2\sqrt{3}}\left(\frac{1}{x-\sqrt{3}} -\frac{1}{x+\sqrt{3}}\right)\right]\mathrm{d} x\\
& =\frac{1}{2\sqrt{2}}\ln\left| \frac{x-\sqrt{2}}{x+\sqrt{2}}\right| +\frac{1}{2\sqrt{3}}\ln\left| \frac{x-\sqrt{3}}{x+\sqrt{3}}\right| +C
\end{aligned}
$$
4. $\displaystyle \int \frac{\mathrm{d} x}{( x-1)^{2}( x-2)}$
解：
$$
\begin{aligned}
& =\int \left[\frac{-1}{( x-1)^{2}} +\frac{1}{x-1} -\frac{1}{x-2}\right]\mathrm{d} x\\
& =\ln\left| \frac{x-1}{x-2}\right| +\frac{1}{x-1} +C
\end{aligned}
$$
5. $\displaystyle \int \frac{x^{2}}{1-x^{4}}\mathrm{d} x$
解：
$$
\begin{aligned}
& =\frac{1}{2}\int \left(\frac{1}{1-x^{2}} -\frac{1}{1+x^{2}}\right)\mathrm{d} x\\
& =-\frac{1}{2}\arctan x+\frac{1}{2}\int \frac{\mathrm{d} x}{( 1-x)( 1+x)}\\
& =-\frac{1}{2}\arctan x+\frac{1}{4}\int \left(\frac{1}{1-x} +\frac{1}{1+x}\right)\mathrm{d} x\\
& =-\frac{1}{2}\arctan x+\frac{1}{4}\ln\left| \frac{1+x}{1-x}\right| +C
\end{aligned}
$$
6. $\displaystyle \int \frac{\mathrm{d} x}{x^{3} +1}$
解：
$$
\begin{aligned}
& =\int \frac{\mathrm{d} x}{( x+1)\left( x^{2} -x+1\right)}\\
& =\frac{1}{3}\int \left(\frac{-1}{x+1} +\frac{-x+2}{x^{2} -x+1}\right)\mathrm{d} x\\
& =\frac{1}{3}\left[ -\ln| x+1| -\frac{1}{2}\ln\left| x^{2} -x+1\right| +\sqrt{3}\arctan\frac{2x-1}{\sqrt{3}}\right] +C
\end{aligned}
$$
7. $\displaystyle \int \frac{\mathrm{d} x}{x^{4} +1}$
解：
$$
\begin{aligned}
& =\int \frac{\mathrm{d} x}{\left[\left( x-\frac{\sqrt{2}}{2}\right)^{2} +\frac{1}{2}\right]\left[\left( x+\frac{\sqrt{2}}{2}\right)^{2} +\frac{1}{2}\right]}\\
& =\int \frac{\mathrm{d} x}{\left[ x-\sqrt{2} x+1\right]\left[ x+\sqrt{2} x+1\right]}\\
& =\frac{1}{4}\int \left(\frac{2-\sqrt{2} x}{1-\sqrt{2} x+x^{2}} +\frac{2+\sqrt{2} x}{1+\sqrt{2} x+x^{2}}\right)\mathrm{d} x\\
& =\frac{1}{8}\left[ -\sqrt{2}\ln\left| x^{2} -\sqrt{2} x+1\right| +4\arctan\left( 2x-\sqrt{2}\right) +\sqrt{2}\ln\left| x^{2} +\sqrt{2} x+1\right| +4\arctan\left( 2x+\sqrt{2}\right)\right] +C\\
& =\frac{1}{8}\left[ 4\sqrt{2}\arctan\left(\sqrt{2} x-1\right) +4\sqrt{2}\arctan\left(\sqrt{2} x+1\right) +\sqrt{2}\ln\frac{x^{2} +\sqrt{2} x+1}{x^{2} -\sqrt{2} x+1}\right] +C
\end{aligned}
$$
8. $\displaystyle \int \frac{x^{3} +x^{2} +2}{\left( x^{2} +2\right)^{2}}\mathrm{d} x$
解：
$$
\begin{aligned}
& =\int \left[\frac{-2x}{\left( x^{2} +2\right)^{2}} +\frac{x+1}{x^{2} +2}\right]\mathrm{d} x\\
\int \frac{x}{\left( x^{2} +2\right)^{2}}\mathrm{d} x & =\frac{1}{2}\int \frac{\mathrm{d}\left( x^{2} +2\right)}{\left( x^{2} +2\right)^{2}}\\
& =-\frac{1}{2\left( x^{2} +2\right)} +C\\
\int \frac{\mathrm{d} x}{x^{2} +2} & =\frac{\sqrt{2}}{2}\int \frac{\mathrm{d}\left(\frac{x}{\sqrt{2}}\right)}{1+\left(\frac{x}{\sqrt{2}}\right)^{2}}\\
& =\frac{\sqrt{2}}{2}\arctan\frac{x}{\sqrt{2}} +C\\
I_{2} & =\frac{1}{4}\left[\frac{x}{x^{2} +2} +\frac{\sqrt{2}}{2}\arctan\frac{x}{\sqrt{2}}\right] +C\\
\int \frac{x}{x^{2} +2}\mathrm{d} x & =\frac{1}{2}\int \frac{\mathrm{d}\left( x^{2} +2\right)}{x^{2} +2}\\
& =\frac{1}{2}\ln\left( x^{2} +2\right) +C
\end{aligned}
$$
原式 $\displaystyle =\frac{1}{2}\ln\left( x^{2} +2\right) +\frac{1}{x^{2} +2} +\frac{\sqrt{2}}{2}\arctan\frac{x}{\sqrt{2}} +C$。
9. $\displaystyle \int \frac{\mathrm{e}^{x}\mathrm{d} x}{\mathrm{e}^{2x} +3\mathrm{e}^{x} +2}$
解：
$$
\begin{aligned}
& =\int \frac{\mathrm{d}\left(\mathrm{e}^{x}\right)}{\mathrm{e}^{2x} +3\mathrm{e}^{x} +2}\\
& =\int \frac{\mathrm{d} y}{y^{2} +3y+2} ,\quad y=\mathrm{e}^{x}\\
& =\int \frac{\mathrm{d} y}{( y+1)( y+2)}\\
& =\int \left(\frac{1}{y+1} -\frac{1}{y+2}\right)\mathrm{d} y\\
& =\ln\left| \frac{y+1}{y+2}\right| +C\\
& =\ln\frac{\mathrm{e}^{x} +1}{\mathrm{e}^{x} +2} +C
\end{aligned}
$$
10. $\displaystyle \int \frac{\cos x\mathrm{d} x}{\sin^{2} x+\sin x-6}$
解：
$$
\begin{aligned}
& =\int \frac{\mathrm{d}(\sin x)}{\sin^{2} x+\sin x-6}\\
& =\int \frac{\mathrm{d} y}{( y+3)( y-2)} ,\quad y=\sin x\\
& =\frac{1}{5}\int \left(\frac{1}{y-2} -\frac{1}{y+3}\right)\mathrm{d} y\\
& =\frac{1}{5}\ln\left| \frac{y-2}{y+3}\right| +C\\
& =\frac{1}{5}\ln\left| \frac{\sin x-2}{\sin x+3}\right| +C
\end{aligned}
$$
11. $\displaystyle \int \frac{x^{3}\mathrm{d} x}{x^{4} +x^{2} +2}$
解：
$$
\begin{aligned}
& =\frac{1}{2}\int \frac{x^{2}\mathrm{d}\left( x^{2}\right)}{x^{4} +x^{2} +2}\\
& =\frac{1}{2}\int \frac{y\mathrm{d} y}{y^{2} +y+2} ,\quad y=x^{2}\\
& =\frac{1}{4}\ln\left| y^{2} +y+2\right| -\frac{1}{2\sqrt{7}}\arctan\frac{2y+1}{\sqrt{7}} +C\\
& =\frac{1}{4}\ln\left( x^{4} +x^{2} +2\right) -\frac{1}{2\sqrt{7}}\arctan\frac{2x^{2} +1}{\sqrt{7}} +C
\end{aligned}
$$
13. $\displaystyle \int \frac{\mathrm{d} x}{2+\sin x}$
解：
$$
\begin{aligned}
& =\int \frac{\mathrm{d} x}{2+\frac{2\tan\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}}\\
& =\frac{1}{2}\int \frac{\left( 1+t^{2}\right)\mathrm{d}( 2\arctan t)}{1+t+t^{2}} ,\quad t=\tan\frac{x}{2}\\
& =\int \frac{\mathrm{d} t}{1+t+t^{2}}\\
& =\frac{\sqrt{12}}{3}\arctan\frac{2\sqrt{3}}{3}\left( t+\frac{1}{2}\right) +C\\
& =\frac{\sqrt{12}}{3}\arctan\frac{\sqrt{3}}{3}\left( 2\tan\frac{x}{2} +1\right) +C
\end{aligned}
$$
14. $\displaystyle \int \frac{\mathrm{d} x}{1+\sin x+\cos x}$
解：
$$
\begin{aligned}
& =\int \frac{\mathrm{d} x}{1+\frac{2\tan\frac{x}{2}}{1+\tan^{2}\frac{x}{2}} +\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}}\\
& =\int \frac{\left( 1+t^{2}\right)\mathrm{d}( 2\arctan t)}{2+2t} ,\quad t=\tan\frac{x}{2}\\
& =\int \frac{\mathrm{d} t}{1+t}\\
& =\ln| 1+t| +C\\
& =\ln\left| 1+\tan\frac{x}{2}\right| +C
\end{aligned}
$$
15. $\displaystyle \int \cot^{4} x\mathrm{d} x$
解：
$$
\begin{aligned}
& =\int t^{4}\mathrm{d}(\operatorname{arccot} t) ,\quad t=\cot x\\
& =\int \frac{-t^{4}}{1+t^{2}}\mathrm{d} t\\
& =\int \left( -t^{2} +1-\frac{1}{1+t^{2}}\right)\mathrm{d} t\\
& =-\frac{t^{3}}{3} +t+\operatorname{arccot} t+C\\
& =-\frac{\cot^{3} x}{3} +\cot x+x+C
\end{aligned}
$$
16. $\displaystyle \int \sec^{4} x\mathrm{d} x$
解：令 $\displaystyle x=\arctan t$，有
$$
\begin{aligned}
& =\int \sqrt{1+t^{2}}^{4}\mathrm{d}(\arctan t)\\
& =\int \left( 1+t^{2}\right)\mathrm{d} t\\
& =t+\frac{t^{3}}{3} +C\\
& =\tan x+\frac{\tan^{3} x}{3} +C
\end{aligned}
$$
17. $\displaystyle \int \frac{\cos x\mathrm{d} x}{5-3\cos x}$
解：
$$
\begin{aligned}
& =\int \frac{\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}}{5-3\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}}\mathrm{d} x\\
& =\int \frac{1-t^{2}}{2+8t^{2}}\mathrm{d}( 2\arctan t) ,\quad t=\tan\frac{x}{2}\\
& =\int \frac{1-t^{2}}{\left( 1+4t^{2}\right)\left( 1+t^{2}\right)}\mathrm{d} t\\
& =\frac{1}{3}\int \left(\frac{5}{1+4t^{2}} -\frac{2}{1+t^{2}}\right)\mathrm{d} t\\
& =\frac{5}{6}\arctan 2t-\frac{2}{3}\arctan t+C\\
& =\frac{5}{6}\arctan\left( 2\tan\frac{x}{2}\right) -\frac{x}{3} +C
\end{aligned}
$$
19. $\displaystyle \int \sin^{5} x\cos^{2} x\mathrm{d} x$
解：
$$
\begin{aligned}
& =\int \left( 1-\cos^{2} x\right)^{2}\cos^{2} x\mathrm{d}( -\cos x)\\
& =-\int \left( 1-y^{2}\right)^{2} y^{2}\mathrm{d} y,\quad y=\cos x\\
& =-\frac{\cos^{3} x}{3} +\frac{2\cos^{5} x}{5} -\frac{\cos^{7} x}{7} +C
\end{aligned}
$$
20. $\displaystyle \int \sin^{6} x\mathrm{d} x$
解：
$$
\begin{aligned}
I_{n} & =\int \sin^{n} x\mathrm{d} x\\
& =-\int \sin^{n-1} x(\mathrm{d}\cos x)\\
& =-\cos x\sin^{n-1} x+( n-1)\int \cos x\cdotp \sin^{n-2} x\cos x\mathrm{d} x\\
& =-\cos x\sin^{n-1} x+( n-1)\int \left(\sin^{n-2} x-\sin^{n} x\right)\mathrm{d} x\\
& =-\cos x\sin^{n-1} x+( n-1)( I_{n-2} -I_{n})\\
I_{n} & =\frac{1}{n}\left(( n-1) I_{n-2} -\cos x\sin^{n-1} x\right)\\
I_{6} & =\frac{5}{6} I_{4} -\frac{1}{6}\cos x\sin^{5} x\\
& =\frac{15}{24} I_{2} -\frac{1}{6}\cos x\sin^{5} x-\frac{5}{24}\cos x\sin^{3} x\\
& =\frac{15}{48} x-\frac{1}{6}\cos x\sin^{5} x-\frac{5}{24}\cos x\sin^{3} x-\frac{15}{48}\cos x\sin x+C
\end{aligned}
$$
23. $\displaystyle \int \frac{\sin x\cos x}{\sin^{2} x+\cos^{4} x}\mathrm{d} x$
解：
$$
\begin{aligned}
& =\int \frac{\sin x}{\sin^{2} x+\left( 1-\sin^{2} x\right)^{2}}\mathrm{d}(\sin x)\\
& =\frac{1}{2}\int \frac{\mathrm{d}\left(\sin^{2} x\right)}{1-\sin^{2} x+\sin^{4} x}\\
& =\frac{1}{2}\int \frac{\mathrm{d} t}{1-t+t^{2}} ,\quad t=\sin^{2} x\\
& =\frac{1}{\sqrt{3}}\arctan\frac{2t-1}{\sqrt{3}} +C\\
& =\frac{1}{\sqrt{3}}\arctan\frac{2\sin^{2} x-1}{\sqrt{3}} +C\\
& =-\frac{1}{\sqrt{3}}\arctan\frac{\cos 2x}{\sqrt{3}} +C
\end{aligned}
$$
25. $\displaystyle \int \sqrt{\frac{1-x}{1+x}}\mathrm{d} x$
解：
$$
\begin{aligned}
& =\int \sqrt{\frac{( 1-x)^{2}}{1-x^{2}}}\mathrm{d} x\\
& =\int \frac{| 1-x| }{\sqrt{1-x^{2}}}\mathrm{d} x\\
& =\int \frac{1-x}{\sqrt{1-x^{2}}}\mathrm{d} x\\
& =\arcsin x-\frac{1}{2}\int \frac{\mathrm{d}\left( x^{2}\right)}{\sqrt{1-x^{2}}}\\
& =\arctan x+\sqrt{1-x^{2}} +C
\end{aligned}
$$
26. $\displaystyle \int \frac{1-\sqrt{x-1}}{1+\sqrt[3]{x-1}}\mathrm{d} x$
解：令 $\displaystyle t=\sqrt[6]{x-1}$
$$
\begin{aligned}
& =\int \frac{1-t^{3}}{1+t^{2}}\mathrm{d}\left( t^{6} +1\right)\\
& =6\int \frac{t^{5} -t^{8}}{1+t^{2}}\mathrm{d} t\\
& =6\int \left( -t^{6} +t^{4} +t^{3} -t^{2} -t+1+\frac{t-1}{1+t^{2}}\right)\mathrm{d} t\\
& =6\left( -\frac{t^{7}}{7} +\frac{t^{5}}{5} +\frac{t^{4}}{4} -\frac{t^{3}}{3} -\frac{t^{2}}{2} +t-\arctan t+\frac{1}{2}\int \frac{\mathrm{d}\left( t^{2}\right)}{1+t^{2}}\right)\\
& =6\left[ -\frac{t^{7}}{7} +\frac{t^{5}}{5} +\frac{t^{4}}{4} -\frac{t^{3}}{3} -\frac{t^{2}}{2} +t-\arctan t+\frac{1}{2}\ln\left( 1+t^{2}\right)\right] +C
\end{aligned}
$$
27. $\displaystyle \int \frac{\sqrt{x+1} +\sqrt{x-1}}{\sqrt{x+1} -\sqrt{x-1}}\mathrm{d} x$
解：
$$
\begin{aligned}
& =\frac{1}{2}\int \left(\sqrt{x+1} -\sqrt{x-1}\right)^{2}\mathrm{d} x\\
& =\int \left( x+\sqrt{x^{2} -1}\right)\mathrm{d} x,\quad x\geq 1\\
I & =\int \sqrt{x^{2} -1}\mathrm{d} x\\
& =x\sqrt{x^{2} -1} -\int x\mathrm{d}\left(\sqrt{x^{2} -1}\right)\\
& =x\sqrt{x^{2} -1} -\int \frac{\left( x^{2} -1\right) +1}{\sqrt{x^{2} -1}}\mathrm{d} x\\
& =x\sqrt{x^{2} -1} -\ln\left( x+\sqrt{x^{2} -1}\right) -I\\
I & =\frac{1}{2}\left[ x\sqrt{x^{2} -1} -\ln\left( x+\sqrt{x^{2} -1}\right)\right] +C
\end{aligned}
$$
因此原式 $\displaystyle =\frac{1}{2}\left[ x^{2} +x\sqrt{x^{2} -1} -\ln\left( x+\sqrt{x^{2} -1}\right)\right] +C$
28. $\displaystyle \int \frac{\mathrm{d} x}{\sqrt[3]{( x+1)^{2}( x-1)^{4}}}$
解：
$$
\begin{aligned}
& =\int \frac{\sqrt[3]{x+1}\mathrm{d} x}{( x-1)( x+1)\sqrt[3]{x-1}}\\
& =\int \frac{1}{x^{2} -1} \cdot \sqrt[3]{\frac{x+1}{x-1}}\mathrm{d} x\\
& =\int \frac{1}{\left(\frac{t+1}{t-1}\right)^{2} -1} \cdotp \sqrt[3]{t}\mathrm{d}\left(\frac{t+1}{t-1}\right) ,\quad t=\frac{x+1}{x-1}\\
& =\int \frac{( t-1)^{2}}{4t}\sqrt[3]{t} \cdot \frac{-2}{( t-1)^{2}}\mathrm{d} t\\
& =-\frac{3}{2}\int \mathrm{d}\left( t^{1/3}\right)\\
& =-\frac{3}{2} t^{1/3} +C\\
& =-\frac{3}{2}\sqrt[3]{\frac{x+1}{x-1}} +C
\end{aligned}
$$
30. $\displaystyle \int \frac{x}{\left( 1+x^{1/3}\right)^{1/2}}\mathrm{d} x$
解：令 $\displaystyle t=\left( 1+x^{1/3}\right)^{1/2}$，有
$$
\begin{aligned}
& =\int \frac{\left( t^{2} -1\right)^{3}}{t}\mathrm{d}\left[\left( t^{2} -1\right)^{3}\right]\\
& =\int \frac{\left( t^{2} -1\right)^{3}}{t} \cdot 3\left( t^{2} -1\right)^{2} \cdot 2t\mathrm{d} t\\
& =6\int \left( t^{2} -1\right)^{5}\mathrm{d} t\\
& =6\int \left( t^{10} -5t^{8} +10t^{6} -10t^{4} +5t^{2} -1\right)\mathrm{d} t\\
& =6\left(\frac{t^{11}}{11} -\frac{5t^{9}}{9} +\frac{10t^{7}}{7} -2t^{5} +2t^{3} -t\right) +C
\end{aligned}
$$
31. $\displaystyle \int \frac{\sqrt{x}}{x^{3/4} +1}\mathrm{d} x$
解：令 $\displaystyle t=x^{1/4}$，有
$$
\begin{aligned}
& =\int \frac{t^{2}}{t^{3} +1}\mathrm{d}\left( t^{4}\right)\\
& =4\int \frac{t^{5}}{t^{3} +1}\mathrm{d} t\\
& =4\int \left( t^{2} -\frac{t^{2}}{t^{3} +1}\right)\mathrm{d} t\\
& =\frac{4}{3} t^{3} -\frac{4}{3}\int \frac{\mathrm{d}\left( t^{3}\right)}{t^{3} +1}\\
& =\frac{4}{3} t^{3} -\frac{4}{3}\ln\left| t^{3} +1\right| +C\\
& =\frac{4}{3} x^{3/4} -\frac{4}{3}\ln\left( x^{3/4} +1\right) +C
\end{aligned}
$$