$\displaystyle \int\frac{1}{e^x+1}dx ($let $\displaystyle u=e^x+1)$

$\displaystyle \int\frac{1}{e^x+1}dx=\int\frac{1}{u}\frac{1}{u-1}du$

$\displaystyle =\int\frac{1}{u^2-u}du$

$\displaystyle =\int\frac{1}{(u-\frac{1}{2})^2}-\frac{1}{4}du$

$\displaystyle =ln(\frac{u-1}{u})+c$

$\displaystyle =ln(\frac{e^x}{e^x+1})+c$