Answer
$2e^{x/2}-2e^{-x/2}+C$
Work Step by Step
$I=\displaystyle \int(e^{x/2}+e^{-x/2})dx=\int e^{x/2}dx+\int e^{-x/2}dx$
Substitutions:
$\left[\begin{array}{lll}
u=x/2 & & \\
du=dx/2 & \Rightarrow & dx=2du
\end{array}\right],\quad \left[\begin{array}{lll}
t=-x/2 & & \\
dt=-dx/2 & \Rightarrow & dx=-2dt
\end{array}\right]$
$I=\displaystyle \int e^{u}(2du)+\int e^{t}(-2dt)$
$= 2e^{u}-2e^{t}+C$
bring back the variable $x$
= $2e^{x/2}-2e^{-x/2}+C$
