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