It's tricky, but 'parts' is probably the best way.
Try:
\(\displaystyle \L\\u=sin^{-1}(2x); \;\ dv=xdx; \;\ v=\frac{x^{2}}{2}; \;\ du=\frac{2}{\sqrt{1-4x^{2}}}\)
\(\displaystyle \L\\\frac{1}{2}x^{2}sin^{-1}(2x)-\int\frac{x^{2}}{\sqrt{1-4x^{2}}}dx\)
There are different aproaches, but you could let:
\(\displaystyle \L\\x=\frac{sin(u)}{2} \;\ and \;\ dx=\frac{cos(u)}{2}du\)
Which will whittle you down to:
\(\displaystyle \L\\\frac{1}{8}\int{sin^{2}(u)}du=\frac{u-sin(u)cos(u)}{16}\)
Resub:
\(\displaystyle \L\\\frac{sin^{-1}(2x)-2x\sqrt{1-4x^{2}}}{16}\)
Don't forget the other portion:
\(\displaystyle \L\\\frac{1}{2}x^{2}sin^{-1}(2x)-\left(\frac{sin^{-1}(2x)-2x\sqrt{1-4x^{2}}}{16}\right)\)
=\(\displaystyle \H\\\left[\frac{x^{2}}{2}-\frac{1}{16}\right]sin^{-1}(2x)+\frac{x\sqrt{1-4x^{2}}}{8}\)
