Answer
$2\sqrt{1+e^{x}}+\ln(\sqrt{1+e^{x}}-1)-\ln(\sqrt{1+e^{x}}+1)+C$
Work Step by Step
Substituting so that $u^{2}=1+e^{x},\qquad (u=\sqrt{1+e^{x}})$
$2udu=e^{x}dx$
$2udu=(u^{2}-1)dx$
$dx=\displaystyle \frac{2udu}{u^{2}-1}$
Since we have a rational function integrand, we can approach with partial fractions.
$\displaystyle \int\frac{2u^{2}du}{u^{2}-1}=\int\frac{2u^{2}-2+2}{u^{2}-1}du=\int 2+\frac{2}{u^{2}-1}$
$\displaystyle \frac{2}{u^{2}-1}$=$\displaystyle \frac{2}{(u-1)(u+1)}$
$\displaystyle \frac{2}{(u-1)(u+1)}=\frac{A}{(u-1)}+\frac{B}{(u+1)}$
$2=Au+A+Bu-B$
$2=(A+B)u+(A-B)\Rightarrow\left\{\begin{array}{ll}
A+B=0 & \\
A-B=2 & \Rightarrow A=1, B=-1
\end{array}\right.$
$\displaystyle \int\frac{2u^{2}du}{u^{2}-1}=\int 2du+\int\frac{1}{u-1}du-\int\frac{1}{u+1}du$
... use table integrals 1 and 2
$=2u+\ln|u-1|-\ln|u+1|+C$
... bring back x ...
$=2\sqrt{1+e^{x}}+\ln|\sqrt{1+e^{x}}-1|-\ln|\sqrt{1+e^{x}}+1|+C$
... the expressions in the absolute value brackets are always positive ...
$=2\sqrt{1+e^{x}}+\ln(\sqrt{1+e^{x}}-1)-\ln(\sqrt{1+e^{x}}+1)+C$
