Answer
$$\int{\sqrt{1+e^x}\ dx}=2\sqrt{1+e^x}-2\ln \left| \frac{\sqrt{1+e^x}+1}{\sqrt{e^x}} \right|+C$$
Work Step by Step
$\displaystyle I=\int{\sqrt{1+e^x}\ dx}$
$\displaystyle{\left[\begin{matrix}
u=\sqrt{1+e^x}& dx=\frac{2\sqrt{1+e^x}\ du}{e^x}& u^2=1+e^x\\
\end{matrix}\right]}$
$I=\displaystyle\int{\sqrt{1+e^x}}\times\frac{2\sqrt{1+e^x}\,\,du}{e^x}\\
I=\displaystyle2\int{\frac{u^2}{u^2-1}du}\\
I=\displaystyle2\int{1+\frac{1}{u^2-1}du}\\
I=\displaystyle2\int{1}\ du+2\int{\frac{1}{u^2-1}du}\\
I=\displaystyle\color{limegreen}{2\int{1}\ du}\color{orangered}{+2\int{\frac{1}{u^2-1}du}}$
$\color{limegreen}{\displaystyle2\int{1\ du}}=\color{limegreen}{2u}$
$\color{orangered}{\displaystyle2\int{\frac{1}{u^2-1}du}}$
$\displaystyle{\left[\begin{matrix}
u=\sec \theta& u^2=\sec ^2\theta& du=\sec \theta \tan \theta d\theta\\
\end{matrix}\right]}$
$\displaystyle2\int{\frac{1}{\sec ^2\theta -1}\sec \theta \tan \theta d\theta}\\
\displaystyle2\int{\frac{1}{\tan ^2\theta}\sec \theta \tan \theta d\theta}\\
\displaystyle2\int{\frac{\sec \theta}{\tan \theta}d\theta}\\
\displaystyle2\int{\csc\theta \ d\theta}\\
\displaystyle2\left( -\ln \left| \csc\theta +\cot \theta \right| \right) \\
\displaystyle\color{orangered}{-2\ln \left| \csc\theta +\cot \theta \right|}$
$I=\displaystyle\color{limegreen}{2u}\color{orangered}{-2\ln \left| \csc\theta +\cot \theta \right|}+C\\
I=\displaystyle2u-2\ln \left| \csc\theta +\cot \theta \right|+C$
$\begin{matrix}
\cot \theta =\frac{1}{\sqrt{u^2-1}}& \csc\theta =\frac{u}{\sqrt{u^2-1}}\\
\end{matrix}$
$\displaystyle I=2u-2\ln \left| \frac{u}{\sqrt{u^2-1}}+\frac{1}{\sqrt{u^2-1}} \right|+C\\
\displaystyle I=2u-2\ln \left| \frac{u+1}{\sqrt{u^2-1}} \right|+C\\
\displaystyle I=2\sqrt{1+e^x}-2\ln \left| \frac{\sqrt{1+e^x}+1}{\sqrt{e^x}} \right|+C$
