Question

Find the value of $\int \frac{1}{e^x+1}dx$.

Answer 1

Let $I=\int \frac{1}{e^x+1}dx$

$I=\int \frac{1}{e^x(1+\frac{1}{e^x})}dx$

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

Let ${1+e}^{-x}=t$

differentiate with respect to $x$

$\frac{d}{dx}\left({1+e}^{-x}\right)\frac{dt}{dx}$

$0-e^{-x}=\frac{dt}{dx}$

$-e^{-x}dx=dt$

$e^{-x}dx=-dt$

$I=-\int \frac{dt}{t}$

$I=-\log t+c$

$I=-\log \left({1+e}^{-x}\right)+c$