Answer
$$\frac{{{x^2}}}{m}\cosh mx - \frac{{2x}}{{{m^2}}}\sinh mx + \frac{2}{{{m^3}}}\cosh mx + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}\sinh mx} dx \cr
& {\text{Integrate by parts, }} \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx,{\text{ }}dv = \sinh mxdx,{\text{ }}v = \frac{1}{m}\cosh mx \cr
& {\text{Using }}\int {udv} = uv - \int {vdu} \cr
& \int {{x^2}\sinh mx} dx = {x^2}\left( {\frac{1}{m}\cosh mx} \right) - \int {\left( {\frac{1}{m}\cosh mx} \right)\left( {2x} \right)} dx \cr
& \int {{x^2}\sinh mx} dx = \frac{{{x^2}}}{m}\cosh mx - \frac{2}{m}\int {x\cosh mx} dx \cr
& {\text{Integrate by parts, }} \cr
& {\text{Let }}u = x,{\text{ }}du = dx,{\text{ }}dv = \cosh mx,{\text{ }}v = \frac{1}{m}\sinh mx \cr
& {\text{Using }}\int {udv} = uv - \int {vdu} \cr
& = \frac{{{x^2}}}{m}\cosh mx - \frac{2}{m}\left( {\frac{x}{m}\sinh mx - \int {\frac{1}{m}\sinh mxdx} } \right) \cr
& = \frac{{{x^2}}}{m}\cosh mx - \frac{{2x}}{{{m^2}}}\sinh mx + \frac{2}{{{m^2}}}\int {\sinh mxdx} \cr
& {\text{Integrating}} \cr
& = \frac{{{x^2}}}{m}\cosh mx - \frac{{2x}}{{{m^2}}}\sinh mx + \frac{2}{{{m^2}}}\left( {\frac{1}{m}\cosh mx} \right) + C \cr
& = \frac{{{x^2}}}{m}\cosh mx - \frac{{2x}}{{{m^2}}}\sinh mx + \frac{2}{{{m^3}}}\cosh mx + C \cr} $$
