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.
 A: $$I = \int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$$
$$I = \int \frac{1}{x^{3} \sqrt{\frac{2}{x}-1}} dx$$
Substitute $\displaystyle u=\frac{1}{x}$ to get
$$I=-\int\frac{u}{\sqrt{2u-1}}du$$
$$-2I=\int\frac{2u-1+1}{\sqrt{2u-1}}du$$
Can you finish it now?
A: 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?
A: If $x<0$ then $\sqrt{2x-x^{2}}$ does not make sense on $\mathbb{R}$ so I suppose that $x>0$, we have
\begin{align*}
\int\frac{1}{x^{2}\sqrt{2x-x^{2}}}{\rm d}x&=\int \frac{1}{x^{2}\sqrt{1-(x-1)^{2}}}{\rm d}x\\&\overset{x\mapsto x-1}{=}\int \frac{1}{(x+1)^{2}\sqrt{1-x^{2}}}{\rm d}x\\&\overset{x\mapsto \sin x}{=}\int \frac{1}{(1+\sin x)^{2}}{\rm d}x\\&\overset{x\mapsto \tan \frac{x}{2}}{=}\int \frac{2}{(x^{2}+1)\left(1+\frac{2x}{x^{2}+1}\right)^{2}}{\rm d}x\\&=2\int \frac{1+x^{2}}{(1+x)^{4}}{\rm d}x\\&=2\int \left(\frac{1}{(x+1)^{2}}-\frac{2}{(x+1)^{3}}+\frac{2}{(x+1)^{4}} \right){\rm d}x\\&=-\frac{2}{3}\left( \frac{3x^{2}+3x+2}{(x+1)^{3}}\right)+C\\&\overset{\text{returning}}{=}-\frac{\sqrt{2x-x^{2}}(x+1)}{3x^{2}}+C
\end{align*}
Details of "returning":
\begin{align*}
-\frac{2}{3}\left( \frac{3x^{2}+3x+2}{(x+1)^{3}}\right)+C&=\frac{\sec^{2}\frac{x}{2}(-3\sin x +\cos x -5)}{3\left(\tan \frac{x}{2}+1\right)^{3}}+C\\
&=\frac{(-3\sin \sin^{-1} x +\cos\sin^{-1}x -5)\sec^{2}\frac{1}{2}\sin^{-1}x)}{3(\tan \frac{1}{2}\sin^{-1}x +1)^{3}}+C\\
&=\frac{2(\sqrt{1-x^{2}}+1)^{2}(\sqrt{1-x^{2}}-3x-5)}{3(\sqrt{1-x^{2}}+x+1)^{3}}+C\\&=\frac{2(\sqrt{-(x-2)x}+1)^{2}(-3x+\sqrt{-(x-2)x}-2)}{3(x+\sqrt{-(x-2)x})^{3}}+C\\&=-\frac{\sqrt{2x-x^{2}}(x+1)}{3x^{2}}+C
\end{align*}
