Answer
$$\frac{{{x^2}}}{2}\arcsin 2x + \frac{1}{8}x\sqrt {1 - 4{x^2}} - \frac{1}{{16}}\arcsin \left( {2x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {x\arcsin 2x} dx \cr
& {\text{Integrating by parts}} \cr
& u = \arcsin 2x,{\text{ }}du = \frac{2}{{\sqrt {1 - {{\left( {2x} \right)}^2}} }}dx \cr
& dv = xdx,{\text{ }}v = \frac{{{x^2}}}{2} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {x\arcsin 2x} dx = \frac{{{x^2}}}{2}\arcsin 2x - \int {\frac{{{x^2}}}{2}} \frac{2}{{\sqrt {1 - {{\left( {2x} \right)}^2}} }}dx \cr
& = \frac{{{x^2}}}{2}\arcsin 2x - \int {\frac{{{x^2}}}{{\sqrt {1 - 4{x^2}} }}} dx \cr
& = \frac{{{x^2}}}{2}\arcsin 2x - \frac{1}{8}\int {\frac{{{{\left( {2x} \right)}^2}}}{{\sqrt {1 - {{\left( {2x} \right)}^2}} }}\left( 2 \right)} dx \cr
& {\text{By tables }}\int {\frac{{{u^2}}}{{\sqrt {{a^2} - {u^2}} }}du = \frac{1}{2}\left( { - u\sqrt {{a^2} - {u^2}} + {a^2}\arcsin \frac{u}{a}} \right) + C} \cr
& = \frac{{{x^2}}}{2}\arcsin 2x - \frac{1}{{16}}\left( { - 2x\sqrt {1 - {{\left( {2x} \right)}^2}} + \arcsin \left( {2x} \right)} \right) + C \cr
& = \frac{{{x^2}}}{2}\arcsin 2x + \frac{1}{8}x\sqrt {1 - 4{x^2}} - \frac{1}{{16}}\arcsin \left( {2x} \right) + C \cr} $$
