Answer
$2\ln|e^{x/2}-e^{-x/2}|+C$
Work Step by Step
$\displaystyle \int\frac{e^{x/2}+e^{-\mathrm{x}/2}}{e^{x/2}-e^{-x/2}}dx=$
If we substitute the expression in the denominator with $u$
$u=e^{x/2}-e^{-x/2},$
we obtain
$du=\displaystyle \frac{1}{2}(e^{x/2}+e^{-x/2})dx$,
(the numerator of the integrand)*$dx$ is substituted with $2du$
$=\displaystyle \int\frac{2du}{u}=2\int\frac{1}{u}du=$
special case of the power rule, $n=-1,$
$=2\ln|u|+C$
bring back the variable $x$
= $2\ln|e^{x/2}-e^{-x/2}|+C$
