中册 4.1 不定积分计算 第13题

\section*{解题过程：} （1）令 $t=\sqrt{x+2}$ ，则 $$ \int \ln (\sqrt{x+2}) \mathrm{d} x=\int 2 t \ln t \mathrm{~d} t=\int \ln t \mathrm{~d}\left(t^{2}\right)=t^{2} \ln t-\frac{1}{2} t^{2}+C=(x+2)\left(\ln \sqrt{x+2}-\frac{1}{2}\right)+C . $$ （2） $\displaystyle \int \ln \left(x+\sqrt{x^{2}-a^{2}}\right) \mathrm{d} x=x \ln \left(x+\sqrt{x^{2}-a^{2}}\right)-\int \frac{x}{\sqrt{x^{2}-a^{2}}} \mathrm{~d} x=x \ln \left(x+\sqrt{x^{2}-a^{2}}\right)-\sqrt{x^{2}-a^{2}}+C$ ． （3） $\displaystyle \int \ln \left(x+\sqrt{x^{2}+1}\right) \mathrm{d} x=x \ln \left(x+\sqrt{x^{2}+1}\right)-\int \frac{x}{\sqrt{x^{2}+1}} \mathrm{~d} x=x \ln \left(x+\sqrt{x^{2}+1}\right)-\sqrt{x^{2}+1}+C$ ． （4） $\displaystyle \int \frac{x \ln \left(x+\sqrt{1+x^{2}}\right)}{\left(1+x^{2}\right)^{2}} \mathrm{~d} x=-\frac{1}{2} \int \frac{\ln \left(x+\sqrt{1+x^{2}}\right)}{\left(1+x^{2}\right)^{2}} \mathrm{~d}\left(\frac{1}{1+x^{2}}\right)$ $$ =-\frac{1}{2} \frac{\ln \left(x+\sqrt{1+x^{2}}\right)}{1+x^{2}}+\frac{1}{2} \int \frac{1}{1+x^{2}} \cdot \frac{1}{\sqrt{1+x^{2}}} \mathrm{~d} x . $$ 令 $x=\tan \theta$ ，则 $$ \int \frac{1}{1+x^{2}} \cdot \frac{1}{\sqrt{1+x^{2}}} \mathrm{~d} x=\int \cos ^{3} \theta \mathrm{~d}(\tan \theta)=\int \cos \theta \mathrm{d} \theta=\sin \theta+C=\frac{x}{\sqrt{1+x^{2}}}+C . $$ 于是 $$ \int \frac{x \ln \left(x+\sqrt{1+x^{2}}\right)}{\left(1+x^{2}\right)^{2}} \mathrm{~d} x=-\frac{1}{2} \frac{\ln \left(x+\sqrt{1+x^{2}}\right)}{1+x^{2}}+\frac{1}{2} \frac{x}{\sqrt{1+x^{2}}}+C . $$ （5） $\displaystyle \int x \ln \frac{1+x}{1-x} \mathrm{~d} x=\frac{x^{2}}{2} \ln \frac{1+x}{1-x}-\int \frac{x^{2}}{2} \cdot \frac{2}{1-x^{2}} \mathrm{~d} x=\frac{x^{2}}{2} \ln \frac{1+x}{1-x}+\int \frac{x^{2}-1+1}{x^{2}-1} \mathrm{~d} x$ $$ =\frac{x^{2}}{2} \ln \frac{1+x}{1-x}+x+\frac{1}{2} \ln \left|\frac{x-1}{x+1}\right|+C . $$ （6）令 $t=\sqrt{x}$ ，则 $$ \begin{aligned} \int \sqrt{x}(\ln x)^{2} \mathrm{~d} x & =2 \int t^{2}\left(\ln t^{2}\right)^{2} \mathrm{~d} t=\frac{2}{3}\left[t^{3}\left(\ln t^{2}\right)^{2}-4 \int t^{2}\left(\ln t^{2}\right) \mathrm{d} t\right]=\frac{2}{3}\left\{t^{3}\left(\ln t^{2}\right)^{2}-\frac{4}{3}\left[t^{3}\left(\ln t^{2}\right)-2 \int t^{2} \mathrm{~d} t\right]\right\} \\ & =\frac{2}{3}\left\{t^{3}\left(\ln t^{2}\right)^{2}-\frac{4}{3}\left[t^{3}\left(\ln t^{2}\right)-\frac{4}{3} t^{3}\right]\right\}+C \end{aligned} $$