Answer
$$
\int x^{2} \sinh (m x) d x=\frac{1}{m} x^{2} \cosh (m x)-\frac{2}{m^{2}} x \sinh (m x)+\\ +\frac{2}{m^{3}} \cosh (m x)+C
$$
where $C$ is an arbitrary constant.
Work Step by Step
$$
\int x^{2} \sinh (m x) d x
$$
Let
$$
\left[\begin{array}{c}
u=x^{2}, \quad d v=\sinh (m x) d x \\
d u=2 x d x \quad v=\frac{1}{m} \cosh (m x)
\end{array}\right]
$$
Integrating by parts, we get
$$\int x^{2} \sinh (m x) d x=\frac{1}{m} x^{2} \cosh (m x)-\frac{2}{m} \int x \cosh (m x) d x
$$
The integral that we obtained, $ \int x \cosh (m x) d x $t, is simpler than the original integral but is still not obvious. Therefore we use integration by parts a second time, this time with
$$
\left[\begin{array}{cc}
U=x, & d V=\cosh (m x) d x, \\
d U=d x & V=\frac{1}{m} \sinh (m x)
\end{array}\right]
$$
Therefor, the original integral can be obtained as the following:
$$
\begin{aligned}
\int x^{2} \sinh (m x) d x&=\frac{1}{m} x^{2} \cosh (m x)-\frac{2}{m} \int x \cosh (m x) d x \\
&=\frac{1}{m} x^{2} \cosh (m x)-\frac{2}{m}\left(\frac{1}{m} x \sinh (m x)-\frac{1}{m} \int \sinh (m x) d x\right)\\
&=\frac{1}{m} x^{2} \cosh (m x)-\frac{2}{m^{2}} x \sinh (m x)+\frac{2}{m^{3}} \cosh (m x)+C
\end{aligned}
$$
where $C$ is an arbitrary constant.