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

Ok, so building off of what lab bhattacharjee said:
$$\int\frac{x^2}{\sqrt{2x-x^2}}dx$$
$$=-\int\sqrt{1-(x-1)^2}dx-\int\dfrac{2-2x}{\sqrt{2x-x^2}}dx+2\int\dfrac1{\sqrt{1-(x-1)^2}}dx$$
Ok, so I use #8 on the 1st integral, u-substitution on the 2nd, and #1 on the 3rd.
$$=-(\frac{(x-1)\sqrt{1-(x-1)^2}}{2}+\frac{1}{2}\arcsin(x-1))-2\sqrt{2x-x^2}+2\arcsin(x-1)+C$$
Simplify.
$$=-\frac{(x-1)\sqrt{1-(x-1)^2}}{2}-2\sqrt{2x-x^2}+\frac{3}{2}\arcsin(x-1)+C$$
