Answer
$\frac{1}{2}e^{2x}+2x-\frac{1}{2}e^{-2x}+C$
Work Step by Step
$\int(e^x+e^{-x})^2dx$
$=\int((e^x)^2+2e^xe^{-x}+(e^{-x})^2)dx$
$=\int(e^{2x}+2+e^{-2x})dx$
$=\int e^{2x}dx+\int 2 dx+\int e^{-2x}dx$
For the first integral, let $u=2x$. Then $du=2dx$, and $\frac{1}{2}du=dx$.
For the third integral, let $v=-2x$. Then $dv=-2dx$, and $-\frac{1}{2}dv=dx$.
$=\int e^u*\frac{1}{2}du+2x+C+\int e^v*(-\frac{1}{2})dv$
$=\frac{1}{2}e^u+2x-\frac{1}{2}e^v+C$
$=\frac{1}{2}e^{2x}+2x-\frac{1}{2}e^{-2x}+C$
