## Calculus 8th Edition

$\frac{1}{2}e^{2x}+2x-\frac{1}{2}e^{-2x}+C$
$\int(e^x+e^{-x})^2dx$ $=\int((e^x)^2+2e^xe^{-x}+(e^{-x})^2)dx$ $=\int(e^{2x}+2+e^{-2x})dx$ $=\int e^{2x}dx+\int 2 dx+\int e^{-2x}dx$ 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 $-\frac{1}{2}dv=dx$. $=\int e^u*\frac{1}{2}du+2x+C+\int e^v*(-\frac{1}{2})dv$ $=\frac{1}{2}e^u+2x-\frac{1}{2}e^v+C$ $=\frac{1}{2}e^{2x}+2x-\frac{1}{2}e^{-2x}+C$