问题及解答

求不定积分

\[
\int\sqrt{\frac{e^x-1}{e^x+1}}dx.
\]

令 $t=\sqrt{\frac{e^x-1}{e^x+1}}$, 则可得

\[
e^x=\frac{1+t^2}{1-t^2},
\]

从而

\[
x=\ln\frac{1+t^2}{1-t^2},\quad dx=\frac{1-t^2}{1+t^2}\cdot\biggl(\frac{1+t^2}{1-t^2}\biggr)'dt=\frac{4t}{(1+t^2)(1-t^2)}dt.
\]

故原积分为

\[
\begin{split}
=&\int t\cdot\frac{4t}{(1+t^2)(1-t^2)}dt\\
=&\int\frac{4(1+t^2)-4}{(1+t^2)(1-t^2)}dt\\
=&4\int\frac{1}{1-t^2}dt+4\int\frac{1}{t^4-1}dt,
\end{split}
\]

而

\[
\int\frac{1}{1-t^2}dt=\frac{1}{2}\int(\frac{1}{1-t}+\frac{1}{1+t})dt=\frac{1}{2}\ln\biggl|\frac{t+1}{t-1}\biggr|+C,
\]

\[
\int\frac{1}{t^4-1}dt=\frac{1}{2}\int(\frac{1}{t^2-1}-\frac{1}{t^2+1})dt=\frac{1}{4}\ln\biggl|\frac{t-1}{t+1}\biggr|-\frac{1}{2}\arctan t+C.
\]

 因此原积分

\[
\begin{split}
=&2\ln\biggl|\frac{t+1}{t-1}\biggr|+\ln\biggl|\frac{t-1}{t+1}\biggr|-2\arctan t+C\\
=&\ln\biggl|\frac{t+1}{t-1}\biggr|-2\arctan t+C\\
=&\ln(e^x+\sqrt{e^{2x}-1})-2\arctan\sqrt{\frac{e^x-1}{e^x+1}}+C.
\end{split}
\]