Answer
$$\frac{{{x^2}}}{2}\arcsin x - \frac{1}{4}\arcsin x + \frac{1}{4}x\sqrt {1 - {x^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {x\arcsin } xdx \cr
& {\text{Integrating by parts,}} \cr
& {\text{Let }}u = \arcsin x,{\text{ }}dv = xdx,{\text{ }}v = \frac{{{x^2}}}{2} \cr
& du = \frac{d}{{dx}}\left[ {\arcsin x} \right] = \frac{1}{{\sqrt {1 - {x^2}} }}dx \cr
& {\text{Using the integration by parts formula}} \cr
& \int {x\arcsin } xdx = \frac{{{x^2}}}{2}\arcsin x - \int {\frac{{{x^2}}}{2}\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)} dx \cr
& = \frac{{{x^2}}}{2}\arcsin x - \frac{1}{2}\int {\frac{{{x^2}}}{{\sqrt {1 - {x^2}} }}} dx{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& {\text{*Integrating }} - \frac{1}{2}\int {\frac{{{x^2}}}{{\sqrt {1 - {x^2}} }}} dx \cr
& {\text{Using the substitution }}x = \sin \theta ,{\text{ }}dx = \cos \theta d\theta \cr
& - \frac{1}{2}\int {\frac{{{x^2}}}{{\sqrt {1 - {x^2}} }}} dx = - \frac{1}{2}\int {\frac{{{{\sin }^2}\theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}\left( {\cos \theta } \right)} d\theta \cr
& = - \frac{1}{2}\int {\frac{{{{\sin }^2}\theta }}{{\sqrt {{{\cos }^2}\theta } }}\left( {\cos \theta } \right)} d\theta \cr
& = - \frac{1}{2}\int {{{\sin }^2}\theta } d\theta \cr
& = - \frac{1}{2}\int {\left( {\frac{1}{2} - \frac{1}{2}\cos 2\theta } \right)} d\theta \cr
& {\text{Integrating}} \cr
& = - \frac{1}{2}\left( {\frac{1}{2}\theta - \frac{1}{4}\sin 2\theta } \right) + C \cr
& = - \frac{1}{2}\left( {\frac{1}{2}\theta - \frac{1}{2}\sin \theta \cos \theta } \right) + C \cr
& = - \frac{1}{4}\theta + \frac{1}{4}\sin \theta \cos \theta + C \cr
& {\text{We know that }}\sin \theta = x,{\text{ then cos}}\theta = \sqrt {1 - {x^2}} \cr
& = - \frac{1}{4}\arcsin x + \frac{1}{4}\left( x \right)\left( {\sqrt {1 - {x^2}} } \right) + C \cr
& = - \frac{1}{4}\arcsin x + \frac{1}{4}x\sqrt {1 - {x^2}} + C \cr
& {\text{Substituting into }}\left( {\bf{1}} \right) \cr
& = \frac{{{x^2}}}{2}\arcsin x - \frac{1}{4}\arcsin x + \frac{1}{4}x\sqrt {1 - {x^2}} + C \cr} $$
