Answer
\begin{aligned}
\int x\ \sinh x\ d x= x \cosh x- \sinh x +C
\end{aligned}
Work Step by Step
Given $$
\int x\ \ \sinh x\ d x
$$
Use integration by parts:
$$ u= x \Rightarrow du= dx $$ $$ dv=\sinh x \ dx \Rightarrow v= \cosh x $$ So, we get
\begin{aligned}
I&=\int x\ \sinh x\ d x\\
&=uv - \int vdu\\
&= x \cosh x-\int \cosh x dx\\
&= x \cosh x- \sinh x +C
\end{aligned}
