How to find the indefinite integral? $$\int\frac{x^2}{\sqrt{2x-x^2}}dx$$
This is the farthest I've got:
$$=\int\frac{x^2}{\sqrt{1-(x-1)^2}}dx$$

As $0<x<2,$
$$\dfrac{x^2}{\sqrt{2x-x^2}}=\dfrac{x^{3/2}}{\sqrt{2-x}}$$
set $x=2\sin^2t,x^{3/2}=\text{?}$
$dx=\text{?}$ and $\sqrt{2-x}=+\sqrt2\cos t$
