Answer
$$ \frac{1}{a}x\sinh ax- \frac{1}{a^2}\cosh ax+c$$
Work Step by Step
Given $$\int x\cosh ax dx $$
Let
\begin{align*}
u&=x\ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=\cosh ax\\
u&= dx\ \ \ \ \ \ \ \ \ \ \ \ \ dv= \frac{1}{a}\sinh ax
\end{align*}
Then using integration by parts
\begin{align*}
\int x\cosh ax dx &=uv-\int vdu\\
&= \frac{1}{a}x\sinh ax- \frac{1}{a}\int \sinh axdx\\
&= \frac{1}{a}x\sinh ax- \frac{1}{a^2}\cosh ax+c
\end{align*}
