Answer
$\displaystyle \frac{\cosh^{3}(x-1)}{3}+C$
Work Step by Step
Use Th.5.18, Derivatives and Integrals of Hyperbolic Functions
$[\cosh u]^{\prime}=(\sinh u)u^{\prime}$
----
$\displaystyle \int\cosh^{2}(x-1)\sinh(x-1)dx=\left[\begin{array}{lll}
u=\cosh(x-1) & & \\
du=\sinh(x-1)dx. & &
\end{array}\right]$
$=\displaystyle \int\underbrace{(\cosh^{2}(x-1))}_{u^{2}}\cdot\underbrace{(\sinh(x-1)dx}_{du}$
$=\displaystyle \frac{u^{3}}{3}+C$
$=\displaystyle \frac{\cosh^{3}(x-1)}{3}+C$
