I was looking for the integral of $\frac{dx}{x^2 \sqrt{2x - x^2}}$
My work:
I was using the reciprocal substitution, where $u = \frac{1}{z}$ and $du = \frac{-1}{z^2}$. Then, I let $x = \frac{1}{z}$ and $dx = \frac{-1}{z^2} dz$, because $u = x$ and $du = dx$.
Substituting these expressions to the integral above, we get:
$$\int \frac{dx}{x^2 \sqrt{2x - x^2}} = \int \frac{ \frac{-1}{z^2} }{ \left (\frac{1}{z} \right)^2 \sqrt{2 \left(\frac{1}{z} \right) - \left ( \frac{1}{z^2} \right)^2}} dz$$ $$ = \int \frac{ \frac{-1}{z^2} }{ \left (\frac{1}{z} \right)^2 \sqrt{ \left(\frac{2}{z} \right) - \left ( \frac{1}{z^2} \right)^2}} dz$$ $$ = \int \frac{ \frac{-1}{z^2} }{ \left (\frac{1}{z} \right)^2 \sqrt{ \left(\frac{2}{z} \right) - \left ( \frac{1}{z^4} \right)}} dz $$ $$ = \int \frac{ -dz }{ \sqrt{\frac{2z^4 z }{z^5} } } $$ $$ = \int \frac{ -dz }{\frac{\sqrt{2z^4 - z }}{\sqrt{z^4 z}}} $$ $$ = \int \frac{ -dz }{\frac{\sqrt{2z^4 - z }}{z^2 z^{\frac{1}{2}}}} $$ $$ = \int \frac{-z^2 dz }{\frac{\sqrt{2z^4 - z }}{z^{\frac{1}{2}}}} $$ $$ = \int \frac{-z^{\frac{5}{2}} dz }{\sqrt{2z^4 - z }} $$ $$ = \int \frac{-z^{\frac{5}{2}} dz }{\sqrt{z(2z^3 - 1)}} $$ $$ = \int \frac{-z^{\frac{5}{2}} dz }{z^{\frac{1}{2}}\sqrt{(2z^3 - 1)}} $$ $$ = \int \frac{-z^2 dz }{\sqrt{(2z^3 - 1)}} $$
Letting $u = 2z^3 - 1 $ and $du = 6z^2 dz$ or $\frac{-du}{6} = -z^2 dz $ we see that...
$$\int \frac{-z^2 dz }{\sqrt{(2z^3 - 1)}} = \int \frac{\frac{du}{-6}}{\sqrt{u}}$$ $$ = \frac{-1}{6} \int u^{\frac{-1}{2}} du$$ $$ = \frac{-1}{6} (2) u^{\frac{1}{2}} du$$ $$ = \frac{-1}{3} u^{\frac{1}{2}} du$$
Then returning back to the original substitution (from $u$ to $z$):
$$ \frac{-1}{3} u^{\frac{1}{2}} du = \frac{-1}{3} (2z^3 - 1)^{\frac{1}{2}} 6z^2 dz$$
Then returning back to the original substitution (from $z$ to $x$):
$$\frac{-1}{3} (2z^3 - 1)^{\frac{1}{2}} 6z^2 dz = -2 (2z^3 - 1)^{\frac{1}{2}}z^2 dz$$
Then.... $$ -2 (2z^3 - 1)^{\frac{1}{2}}z^2 dz = = -2 \left(2\left( \frac{1}{x}\right)^3 - 1\right)^{\frac{1}{2}}\left( \frac{1}{x}\right)^2 \left( \frac{1}{x}\right)^2 dx $$ $$ = -2 \left(2\left( \frac{1}{x^3}\right) - 1\right)^{\frac{1}{2}}\left( \frac{1}{x^4}\right) dx $$ $$ = -2 \left(\left( \frac{2}{x^3}\right) -1\right)^{\frac{1}{2}}\left( \frac{1}{x^4}\right) dx $$ $$ = -2 \left(\left( \frac{2-x^3}{x^3}\right)\right)^{\frac{1}{2}}\left( \frac{1}{x^4}\right) dx $$ $$ = -2 \left(\left( \frac{(2-x^3)^{\frac{1}{2}}}{(x^3)^{\frac{1}{2}}}\right)\right)\left( \frac{1}{x^4}\right) dx $$ $$ = \left( -2 \frac{(2-x^3)^{\frac{1}{2}}}{x^{\frac{3}{2}}}\right)\left( \frac{1}{x^4}\right) dx $$ $$ = \left( -2 \frac{(2-x^3)^{\frac{1}{2}}}{x^{\frac{11}{2}}}\right) dx $$ $$\int \frac{dx}{x^2 \sqrt{2x - x^2}} = -2 (2-x^3)^{\frac{1}{2}}{x^{-\frac{11}{2}}} dx $$
I've done my best, but to no avail.....How to evaluate $\int \frac{dx}{x^2 \sqrt{2x - x^2}}$ properly using reciprocal substitution?
I think that you took a very long way; so, let me try.
Fist, complete the square to write $$I={\displaystyle\int}\dfrac{dx}{x^2\sqrt{2x-x^2}}={\displaystyle\int}\dfrac{dx}{x^2\sqrt{1-\left(x-1\right)^2}}$$ Now $x=u+1$ gives $$I={\displaystyle\int}\dfrac{du}{\left(u+1\right)^2\sqrt{1-u^2}}$$ Now, use $u=\sin(v)$ which gives $$I={\displaystyle\int}\dfrac{\cos\left(v\right)}{\left(\sin\left(v\right)+1\right)^2\sqrt{1-\sin^2\left(v\right)}}\,dv={\displaystyle\int}\dfrac{dv}{\left(\sin\left(v\right)+1\right)^2}$$ Now, use the tangent half-angle substitution $w=\tan\left(\dfrac{v}{2}\right)$ $$I={2}{\displaystyle\int}\dfrac{w^2+1}{\left(w^2+2w+1\right)^2}\,dw=2{\displaystyle\int}\dfrac{w^2+1}{\left(w+1\right)^4}\,dw$$ Now $w=z-1$ makes $$I=2\int \frac{z^2-2z+2}{z^4}\,dz$$ which looks to be simple.