Answer
\[x\sinh x-\cosh x+C\]
Work Step by Step
Let \[I=\int x\cosh x \:dx\]
Using integration by parts
\[I=x\int \cosh x dx-\int\left((x)'\int \cosh x \: dx\right)dx\]
\[ \left[\int\cosh xdx=\sinh x\right]\]
\[I=x\sinh x-\int \sinh x\: dx\]
\[\left[\int\sinh xdx=\cos hx\right]\]
\[I=x\sinh x-\cos hx+C\]
Hence \[I=x\sinh x-\cos hx+C\]
