Question

Evaluate $\int \frac{1}{1+e^{-x}}dx$

Answer 1

Given $I=\int \frac{1}{1+e^{-x}}dx$
$=\int \frac{1}{1+\frac{1}{e^x}}dx$
$=\int \frac{e^x}{1+e^x}dx$
Let $t=1+e^x$, then $dt=e^{x}dx$
$I=\int \frac{1}{t}dt$
$=\ln |t|+C$ [Since, $\int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\ln f\left(x\right)$]
$\therefore I=ln(1+e^x)+C$