Answer
$\ln|\sinh x|+C$
Work Step by Step
Use Th.5.18, Derivatives and Integrals of Hyperbolic Functions
$[\sinh u]^{\prime}=(\cosh u)u^{\prime}$
----
$\displaystyle \int\frac{\cosh x}{\sinh x}dx=\left[\begin{array}{lll}
u=\sinh x & & \\
du=(\cosh x)dx. & &
\end{array}\right]$
$=\displaystyle \int\underbrace{\frac{1}{\sinh x}}_{u^{-1}}\cdot\underbrace{(\cosh x)dx}_{du}$
$=\displaystyle \int \frac{du}{u}$
$=\ln|u|+C$
$=\ln|\sinh x|+C$
