来自群的一道分式根号指数积分

(4.6 KB)

2012-12-15 18:16

令 $u=\sqrt{e^x-1}$，则 $x=\ln (u^2+1)$，故
\begin{align*}
\int\frac{xe^x}{\sqrt{e^x-1}}\rmd x&=\int\frac{(u^2+1)\ln (u^2+1)}u\rmd{\ln (u^2+1)} \\
& =2\int\ln (u^2+1)\rmd u \\
& =2u\ln (u^2+1)-2\int u\rmd{\ln (u^2+1)} \\
& =2u\ln (u^2+1)-4\int\frac{u^2}{u^2+1}\rmd u \\
& =\cdots
\end{align*}