What is the integral $\;\displaystyle\int\left[x^2(x-2)\right]^{-2/3}\text{d}x\;$? What is the integral $\;\displaystyle\int\left[x^2(x-2)\right]^{-2/3}\text{d}x\;$?
I have tried solving this integral by using U-substitution and integration by parts but got nowhere. Don't really know what else to try. WolframAlpha gives an answer to the integral but is unable to show steps for the solution. Any help would be appreciated!
 A: We have
$$ I = \int x^{-\frac43}(x-2)^{-\frac23}dx = \int \frac{1}{x(x-2)}\sqrt[3]{\frac{x-2}{x}} d x$$
Integrals of the form $$\int Q(x) \sqrt[n]{\frac{ax+b}{cx+d}}dx $$ where $Q$ is a rational function can be solved using the substitution
$$ t = \sqrt[n]{\frac{ax+b}{cx+d}} $$
In this case, we have
$$ t = \sqrt[3]{\frac{x-2}{x}}$$
that is
$$ x= \frac{2}{1-t^3}$$
It will transform the integral into an integral of a rational function.
A: Integral-calculator.com gives "steps". I copy below, but not sure if they are "useful"-
Problem:
$$
\int \frac{1}{\left((x-2) x^{2}\right)^{\frac{2}{3}}} d x
$$
Rewrite/simplify:
$$
=\int \frac{1}{(x-2)^{\frac{2}{3}} x^{\frac{4}{3}}} \mathrm{~d} x 
$$ Substitute $u=\sqrt[3]{x} \longrightarrow \frac{\mathrm{d} u}{\mathrm{~d} x}=\frac{1}{3 x^{\frac{2}{3}}} \longrightarrow \mathrm{d} x=3 x^{\frac{2}{3}} \mathrm{~d}u$:
$$\tiny x=u^{3} \\
\tiny \frac1{x^{2/3}} =\frac1{u^2}\\
=3 \int \frac{1}{u^{2}\left(u^{3}-2\right)^{\frac{2}{3}}} \mathrm{~d} u
$$
Substitute $v=\frac{\sqrt[3]{u^{3}-2}}{u} \rightarrow \frac{\mathrm{d} v}{\mathrm{~d} u}=\frac{u}{\left(u^{3}-2\right)^{\frac{2}{3}}}-\frac{\sqrt[3]{u^{3}-2}}{u^{2}} \longrightarrow \mathrm{d} u=\frac{1}{\frac{u}{\left(u^{3}-2\right)^{\frac{2}{3}}}-\frac{\sqrt[3]{u^{3}-2}}{u^{2}}} \cdot \mathrm{d}  v$
$$=\int \frac{1}{2} \mathrm{~d} v$$
(from here it is obvious so I stop)
