Integrate : $\int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$ Integrate :
$$I = \int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$$
My attempt : substitute $\sin t = x-1$, $u = \tan \frac{t}{2}$
$$I = \int \frac{1}{x^{2} \sqrt{1-(x-1)^{2}}} dx = \int \frac{1}{(\sin t + 1)^{2} \cos t} \cos t dt$$
$$= \int \frac{1}{(\sin t + 1)^{2}} dt = \int \frac{2(u^{2}+1)}{(u+1)^{4}} du$$
$$= 2 \int \left(\frac{1}{(u+1)^{2}} - \frac{2}{(u+1)^{3}} + \frac{2}{(u+1)^{4}} \right)$$
$$= - \frac{2}{u+1} + \frac{2}{(u+1)^{2}} - \frac{4}{3(u+1)^{3}} + const.$$
$$= \frac{-2(3u^{2}+3u+2)}{3(u+1)^{3}} + const.$$
where
$$A = \sqrt{2x-x^{2}},~~ u = \sqrt{\frac{1-A}{1+A}}$$
However, Wolfram gives the following answer :
$$ - \frac{\sqrt{2x-x^{2}}~(x+1)}{3x^{2}} + const. $$
which is way more simple than mine. Is there any other way to integrate this? The $u$ in my answer is very complicated, I just couldn't change my answer into Wolfy's.

If $x>0,$
Let $1-x=\cos2t,dx=2\sin2t\ dt$
so we get $$\int\dfrac{dt}{\sin^4t}dt=\int(\cot^2t+1)\csc^2t\ dt$$
Can you take it home from here?
