计算 $F(x)=\int \frac{1}{\sqrt{1+e^{x}}+\sqrt{1-e^{x}}} \mathrm{~d} x$
【参考解答】: 第一步: 有理化
\[F(x)=\int \frac{\sqrt{1+e^{x}}-\sqrt{1-e^{x}}}{2 e^{x}} \mathrm{~d} x\]第二步: 指数换元. 令 $e^{x}=t, x=\ln t, \mathrm{~d} x=\frac{1}{t} \mathrm{~d} t$ ，
\[\begin{aligned} & F(x)=\int \frac{\sqrt{1+t}-\sqrt{1-t}}{2 t^{2}} \mathrm{~d} t \\ =& \int \frac{\sqrt{1+t}}{2 t^{2}} \mathrm{~d} t-\int \frac{\sqrt{1-t}}{2 t^{2}} \mathrm{~d} t=F_{1}(t)+F_{2}(t) \end{aligned}\]第三步: 根式换元. 对第一个积分，令 $\sqrt{1+t}=u$, 得 $t=u^{2}-1, \mathrm{~d} t=2 u \mathrm{~d} u$, 于是
\[F_{1}(t)=\int \frac{u}{2\left(u^{2}-1\right)^{2}} 2 u \mathrm{~d} u=\int \frac{u^{2}}{\left(u^{2}-1\right)^{2}} \mathrm{~d} u\]第四步: 分解部分分式，得
\[\begin{aligned} \frac{u^{2}}{\left(u^{2}-1\right)^{2}}=& \frac{1}{4(u-1)^{2}}-\frac{1}{4(u+1)} \\ &+\frac{1}{4(u+1)^{2}}+\frac{1}{4(u-1)} \end{aligned}\]所以 $F_{1}(t)=\int \frac{u^{2}}{\left(u^{2}-1\right)^{2}} \mathrm{~d} u$
\[=-\frac{1}{4}\left[\frac{1}{u-1}+\ln (u-1)+\frac{1}{u+1}-\ln (u+1)\right]+C\] \[=-\frac{1}{2} \frac{u}{u^{2}-1}-\frac{1}{4} \ln \frac{u+1}{u-1}+C \\=-\frac{1}{2} \frac{\sqrt{1+t}}{t}-\frac{1}{4} \ln \frac{\sqrt{1+t}+1}{\sqrt{1+t}-1}+C\]类似第三步、令 $\sqrt{1-t}=u$, 得
\[F_{2}(t)=\int \frac{u^{2}}{\left(1-u^{2}\right)^{2}} \mathrm{~d} u\]其部分分式与上面一样，可得
\[\begin{aligned} & F_{2}(t)=\int \frac{u^{2}}{\left(1-u^{2}\right)^{2}} \mathrm{~d} u \\ =&-\frac{1}{2} \frac{u}{u^{2}-1}-\frac{1}{4} \ln \frac{u+1}{u-1}+C \\ =& \frac{1}{2} \frac{\sqrt{1-t}}{t}-\frac{1}{4} \ln \frac{\sqrt{1-t}+1}{\sqrt{1-t}-1}+C \end{aligned}\]第五步: 回代 $t=e^{x}$ 得
\[\begin{gathered} F(x)=F_{1}(t)+F_{2}(t) \\ =-\frac{1}{2} \frac{\sqrt{1+t}}{t}-\frac{1}{4} \ln \frac{\sqrt{1+t}+1}{\sqrt{1+t}-1} \\ +\frac{1}{2} \frac{\sqrt{1-t}}{t}-\frac{1}{4} \ln \frac{\sqrt{1-t}+1}{\sqrt{1-t}-1}+C \\ =-\frac{1}{2} \frac{\sqrt{1+e^{x}}}{e^{x}}-\frac{1}{4} \ln \frac{\sqrt{1+e^{x}}+1}{\sqrt{1+e^{x}}-1} \\ +\frac{1}{2} \frac{\sqrt{1-e^{x}}}{e^{x}}-\frac{1}{4} \ln \frac{\sqrt{1-e^{x}}+1}{\sqrt{1-e^{x}}-1}+C \end{gathered}\]