Answer
$\displaystyle \frac{1}{2}(e^{x}-e^{-x})+C$
Work Step by Step
$I= \displaystyle \int\frac{e^{x}+e^{-x}}{2}dx=\int\frac{e^{x}}{2}dx+\int\frac{e^{-x}}{2}dx$
$=\displaystyle \frac{1}{2}\int e^{x}dx+\frac{1}{2}\int e^{-x}dx$
first term: standard integral, $\displaystyle \int e^{x}dx=e^{x}+C$
second term: substitute $\left[\begin{array}{l}
u=-x\\
du=-dx
\end{array}\right]$
$=\displaystyle \frac{e^{x}}{2}+\frac{1}{2}\int u(-du)$
$=\displaystyle \frac{e^{x}}{2}-\frac{1}{2}\int udu$
$=\displaystyle \frac{e^{x}}{2}-\frac{e^{u}}{2}+C$
bring back the variable $x$
= $\displaystyle \frac{1}{2}(e^{x}-e^{-x})+C$
