Answer
$$\frac{{\pi - 2}}{4}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {x\arcsin {x^2}} dx \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = \arcsin {x^2},{\text{ }}du = \frac{{2x}}{{\sqrt {1 - {x^4}} }}dx \cr
& dv = xdx,{\text{ }}v = \frac{{{x^2}}}{2} \cr
& {\text{Integration by parts formula }} \cr
& \int {\underbrace {\arcsin {x^2}}_u} \underbrace {xdx}_{dv} = \underbrace {\left( {\arcsin {x^2}} \right)}_u\underbrace {\left( {\frac{{{x^2}}}{2}} \right)}_v - \int {\underbrace {\left( {\frac{{{x^2}}}{2}} \right)}_v} \underbrace {\left( {\frac{{2x}}{{\sqrt {1 - {x^4}} }}} \right)dx}_{du} \cr
& \int {x\arcsin {x^2}} dx = \frac{{{x^2}}}{2}\arcsin {x^2} - \int {\frac{{{x^3}}}{{\sqrt {1 - {x^4}} }}} dx \cr
& \int {x\arcsin {x^2}} dx = \frac{{{x^2}}}{2}\arcsin {x^2} + \frac{1}{4}\int {\frac{{ - 4{x^3}}}{{\sqrt {1 - {x^4}} }}} dx \cr
& \int {x\arcsin {x^2}} dx = \frac{{{x^2}}}{2}\arcsin {x^2} + \frac{1}{4}\left( {\frac{{\sqrt {1 - {x^4}} }}{{1/2}}} \right) + C \cr
& \int {x\arcsin {x^2}} dx = \frac{{{x^2}}}{2}\arcsin {x^2} + \frac{1}{2}\sqrt {1 - {x^4}} + C \cr
& {\text{Therefore,}} \cr
& \int_0^1 {x\arcsin {x^2}} dx = \left[ {\frac{{{x^2}}}{2}\arcsin {x^2} + \frac{1}{2}\sqrt {1 - {x^4}} } \right]_0^1 \cr
& = \left[ {\frac{{{{\left( 1 \right)}^2}}}{2}\arcsin {{\left( 1 \right)}^2} + \frac{1}{2}\sqrt {1 - {{\left( 1 \right)}^4}} } \right] - \left[ {\frac{{{{\left( 0 \right)}^2}}}{2}\arcsin {{\left( 0 \right)}^2} + \frac{1}{2}\sqrt {1 - {{\left( 0 \right)}^4}} } \right] \cr
& = \left[ {\frac{1}{2}\left( {\frac{\pi }{2}} \right) - 0} \right] - \left[ {0 + \frac{1}{2}} \right] \cr
& = \frac{\pi }{4} - \frac{1}{2} \cr
& = \frac{{\pi - 2}}{4} \cr} $$
