中册 4.1 不定积分计算 第20题

\section*{解题过程：} （1）令 $t=x^{5}$ ，则 $$ \begin{aligned} \int \frac{x^{14}}{\left(1+x^{5}\right)^{4}} \mathrm{~d} x & =\frac{1}{5} \int \frac{x^{10}}{\left(1+x^{5}\right)^{4}} \mathrm{~d}\left(x^{5}\right)=\frac{1}{5} \int \frac{t^{2}}{(1+t)^{4}} \mathrm{~d} t=\frac{1}{5} \int \frac{(t+1)^{2}-2(t+1)+1}{(1+t)^{4}} \mathrm{~d} t \\ & =\frac{1}{5}\left[-\frac{1}{(1+t)}+\frac{1}{(1+t)^{2}}-\frac{1}{3} \frac{1}{(1+t)^{3}}+C\right]=\frac{1}{5}\left[-\frac{1}{\left(1+x^{5}\right)}+\frac{1}{\left(1+x^{5}\right)^{2}}-\frac{1}{3} \frac{1}{\left(1+x^{5}\right)^{3}}+C\right] . \end{aligned} $$ （2）令 $t=x^{3}$ ，则 $$ \begin{aligned} \int \frac{1}{x\left(1+x^{3}\right)^{2}} \mathrm{~d} x & =\int \frac{x^{2}}{x^{3}\left(1+x^{3}\right)^{2}} \mathrm{~d} x=\frac{1}{3} \int \frac{1}{x^{3}\left(1+x^{3}\right)^{2}} \mathrm{~d}\left(x^{3}\right)=\frac{1}{3} \int \frac{1}{t(1+t)^{2}} \mathrm{~d} t=\frac{1}{3} \int\left[\frac{1}{t(1+t)}-\frac{1}{(1+t)^{2}}\right] \mathrm{d} t \\ & =\frac{1}{3}\left(\ln \left|\frac{t}{1+t}\right|+\frac{1}{1+t}\right)+C=\frac{1}{3}\left(\ln \left|\frac{x^{3}}{1+x^{3}}\right|+\frac{1}{1+x^{3}}\right)+C \end{aligned} $$ （3）方法 1： $\displaystyle \int \frac{1}{x\left(x^{n}+1\right)} \mathrm{d} x=\int \frac{1}{x^{n+1}\left(1+x^{-n}\right)} \mathrm{d} x=-\frac{1}{n} \int \frac{\mathrm{~d}\left(1+x^{-n}\right)}{1+x^{-n}}$ $$ =-\frac{1}{n} \ln \left|1+x^{-n}\right|+C=\frac{1}{n} \ln \left|\frac{x^{n}}{1+x^{n}}\right|+C . $$ 方法 2： $\displaystyle \int \frac{1}{x\left(x^{n}+1\right)} \mathrm{d} x=\int\left(\frac{1}{x}-\frac{x^{n-1}}{1+x^{n}}\right) \mathrm{d} x=\ln |x|-\frac{1}{n} \ln \left|1+x^{n}\right|+C$ ． （4） $\displaystyle \int \frac{1}{x^{2}(1+x)} \mathrm{d} x=\int \frac{1-x}{x^{2}} \mathrm{~d} x+\int \frac{1}{1+x} \mathrm{~d} x=-\frac{1}{x}-\ln |x|+\ln |1+x|+C$ ． （5） $\displaystyle \int \frac{\mathrm{d} x}{x^{4}\left(1+x^{2}\right)}=\int\left(\frac{1-x^{2}}{x^{4}}+\frac{1}{1+x^{2}}\right) \mathrm{d} x=\int\left(x^{-4}-x^{-2}\right) \mathrm{d} x+\int \frac{\mathrm{d} x}{1+x^{2}}=-\frac{1}{3} x^{-3}+x^{-1}+\arctan x+C$ ． （6） $\displaystyle \int \frac{4 x^{3}+x^{7}}{x^{8}+9} \mathrm{~d} x=4 \int \frac{x^{3}}{x^{8}+9} \mathrm{~d} x+\int \frac{x^{7}}{x^{8}+9} \mathrm{~d} x=\frac{1}{8} \ln \left(9+x^{8}\right)+\frac{1}{3} \arctan \frac{x^{4}}{3}+C$ ． （7） $\displaystyle \int \frac{x^{2}}{(1-x)^{n}} \mathrm{~d} x=\int \frac{(1-x)^{2}-2(1-x)+1}{(1-x)^{n}} \mathrm{~d} x=\int\left[(1-x)^{2-n}-2(1-x)^{1-n}+(1-x)^{-n}\right] \mathrm{d} x$ $$ =\frac{1}{(n-1)(1-x)^{n-1}}-\frac{2}{(n-1)(1-x)^{n-2}}+\frac{1}{(n-3)(1-x)^{n-3}}+C . $$