How to find the indefinite integral?

Question

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

Answer 1

Hint:

As $\dfrac{d(2x-x^2)}{dx}=2-2x$

$$\dfrac{x^2}{\sqrt{2x-x^2}}=\dfrac{x^2-2x+2x-2+2}{\sqrt{2x-x^2}}$$

$$=-\sqrt{1-(x-1)^2}-\dfrac{2-2x}{\sqrt{2x-x^2}}+\dfrac2{\sqrt{1-(x-1)^2}}$$

Now use $\#1,\#8$ of this

Answer 2

Hint: substitute $$x-1= \sin t $$ so as $$\\ x-1= \sin t \\ x=\sin t +1\\ dx = \cos t\,dt \\ \int \frac { x^2 }{ \sqrt { 1-(x-1)^2 } } \, dx=\int \frac { \cos t (\sin t +1)^2 }{ \sqrt { 1-\sin^2{t} } } \, dt=\int (\sin t +1)^2 \, dt \\ $$

Answer 3

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 $\sqrt{2-x}=+\sqrt2\cos t$

Answer 4

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. $$=-\frac{(x-1)\sqrt{1-(x-1)^2}}{2}-2\sqrt{2x-x^2}+\frac{3}{2}\arcsin(x-1)+C$$