Answer
$$\frac{1}{2}{x^2}{\sin ^{ - 1}}x - \frac{1}{4}{\sin ^{ - 1}}x + \frac{1}{4}x\sqrt {1 - {x^2}} + C$$
Work Step by Step
$$\eqalign{
& \int x {\sin ^{ - 1}}xdx \cr
& {\text{Using the integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}\,\,\,\,\,u = {\sin ^{ - 1}}x,\,\,\,\,du = \frac{1}{{\sqrt {1 - {x^2}} }}dx\, \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = xdx,\,\,\,\,v = \frac{1}{2}{x^2} \cr
& \cr
& {\text{Integration by parts formula }}\cr
& \int {udv} = uv - \int {vdu}\cr
& {\text{Then}}\cr
& \int x {\sin ^{ - 1}}xdx = \frac{1}{2}{x^2}{\sin ^{ - 1}}x - \int {\frac{1}{2}{x^2}} \left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)\,\,\,dx \cr
& \int x {\sin ^{ - 1}}xdx = \frac{1}{2}{x^2}{\sin ^{ - 1}}x - \int {\frac{{{x^2}}}{{2\sqrt {1 - {x^2}} }}dx} \cr
& \cr
& {\text{Integrating }}\int {\frac{{{x^2}}}{{2\sqrt {1 - {x^2}} }}dx} \cr
& {\text{let }}x = \sin \theta ,\,\,\,\,\,dx = \cos \theta d\theta \cr
& \,\,\,\int {\frac{{{x^2}}}{{2\sqrt {1 - {x^2}} }}dx} = \int {\frac{{{{\sin }^2}\theta }}{{2\sqrt {1 - {{\sin }^2}\theta } }}\left( {\cos \theta d\theta } \right)} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int {\frac{{{{\sin }^2}\theta }}{{2\cos \theta }}\left( {\cos \theta d\theta } \right)} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int {\frac{{{{\sin }^2}\theta }}{2}} d\theta \cr
& {\text{Use the identity co}}{{\text{s}}^2}\theta = \frac{1}{2} - \frac{1}{2}\cos 2\theta \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{4}\int {\left( {1 - \cos 2\theta } \right)} d\theta \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{4}\theta - \frac{1}{4}\left( {\frac{1}{2}\sin 2\theta } \right) + C \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{4}\theta - \frac{1}{8}\sin 2\theta + C \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{4}\theta - \frac{1}{4}\sin \theta \cos \theta + C \cr
& \cr
& {\text{Replace }}\theta = {\sin ^{ - 1}}x \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{4}{\sin ^{ - 1}}x - \frac{1}{4}\sin \left( {{{\sin }^{ - 1}}x} \right)\cos \left( {{{\sin }^{ - 1}}x} \right) + C \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{4}{\sin ^{ - 1}}x - \frac{1}{4}x\sqrt {1 - {x^2}} + C \cr
& {\text{Then}}{\text{,}} \cr
& \frac{1}{2}{x^2}{\sin ^{ - 1}}x - \int {\frac{{{x^2}}}{{2\sqrt {1 - {x^2}} }}dx} \cr
& \frac{1}{2}{x^2}{\sin ^{ - 1}}x - \frac{1}{4}{\sin ^{ - 1}}x + \frac{1}{4}x\sqrt {1 - {x^2}} + C \cr} $$
