Answer
$\displaystyle\int{\frac{1}{1+e^x}dx}=\ln \left| \frac{e^x}{e^x+1} \right|+C\\
$
Work Step by Step
$I=\displaystyle\int{\frac{1}{1+e^x}dx}$
$\displaystyle{\begin{matrix}
u=1+e^x& u-1=e^x\\
\frac{du}{dx}=e^x& dx=\frac{du}{e^x}\\
\end{matrix}}$
$I=\displaystyle\int{\frac{1}{u}\frac{du}{u-1}}\\
I=\displaystyle\int{\frac{1}{u\left( u-1 \right)}du}$
$\displaystyle\frac{1}{u\left( u-1 \right)}\equiv \frac{A}{u}+\frac{B}{u-1}$
$1=A\left( u-1 \right) +Bu\\
1=Au-A+Bu\\
1=\left( A+B \right) u-A\\
u=0\ \therefore \ A=-1\\
u=-1\ \therefore \ 1=-1+B+1\\
B=1$
$I=\displaystyle\int{\frac{1}{u-1}du}-\int{\frac{1}{u}du}\\
I=\displaystyle{\ln \left| u-1 \right|-\ln \left| u \right|+C}\\
I=\displaystyle\ln \left| \frac{u-1}{u} \right|+C\\
I=\displaystyle\ln \left| \frac{e^x}{e^x+1} \right|+C\\
$
