Answer
$x\sinh x-\cosh x+C$
Work Step by Step
$\displaystyle \int x\cosh xdx=$ by parts,
$\displaystyle \left[\begin{array}{ll}
u=x & dv=\cosh xdx\\
du=dx & v=\sinh x
\end{array}\right],\quad \int udv=uv-\int vdu$
$=x\displaystyle \sinh x-\int\sinh xdx $ $[ \int sinhxdx = -coshx]$
$=x\sinh x+\cosh x+C$
