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