[(x^2)/(squareroot2x - x^2) ]dx
do I do u-substitution?
$\displaystyle \int\frac{x^{2}}{\sqrt{2x-x^{2}}}dx$
You could let $\displaystyle x=u+1, \;\ dx=du$
Make the subs and get:
$\displaystyle \int\frac{(u+1)^{2}}{\sqrt{1-u^{2}}}du$
Now, we can let $\displaystyle u=sin(t), \;\ du=cos(t)dt$
Make the subs and it whittles down to:
$\displaystyle \int{(1+sin(t))^{2}}dt$
$\displaystyle \int{sin^{2}(t)}dt+2\int{sin(t)}dt+\int{dt}$
$\displaystyle \frac{-1}{2}sin(t)cos(t)-2cos(t)+\frac{3}{2}t$
Resub for $\displaystyle t=sin^{-1}(u)$
That gives us:
$\displaystyle \frac{3sin^{-1}(u)}{2}-\frac{u\sqrt{1-u^{2}}}{2}-2\sqrt{1-u^{2}}$
Resub for $\displaystyle u=x-1$ and get:
$\displaystyle \frac{3}{2}sin^{-1}(x-1)-\frac{x\sqrt{2x-x^{2}}}{2}-\frac{3}{2}\sqrt{2x-x^{2}}$
Ok
we have $\displaystyle \int\frac{x^2}{\sqrt{2x-x^2}}dx$
Completing the square we get
$\displaystyle \int\frac{x^2}{\sqrt{1-(x-1)^2}}$
Now letting $\displaystyle x-1=\sin(\theta)$
so $\displaystyle x=\sin(\theta)-1\Rightarrow{dx=\cos(\theta)}$
Giving us
$\displaystyle \int\frac{(\sin(\theta)-1)^2}{\sqrt{1-\sin^2(\theta)}}d\theta$
Can you go from there?