<s>[INST] How to integrate $((x^2-1)(x+1))^{-2/3}$ using the substitution $u=(x-1)/(x+1)$? I was asked to find the indefinite integral $$\int  \frac{1}{((x^2-1)(x+1))^{2/3}}  dx$$ 
using the substitution of $u=(x-1)/(x+1)$. 
How do I make this substitution?
I attempted to solve this problem as follows: First I factored the denominator and distributed the exponent. Then I multiplied by one in the form of (x-1)/(x-1) to yield $$\int  {(x-1)^{-4/3}((x-1)/(x+1))^{2/3}(x+1)^{-2/3}}  dx$$ At this point I made the u substitution and replace all x with $((-u-1)/(u-1))$ I did some rearranging and got stuck with (1/2) $$\int  u^{2/3}(-2/(u-1))^{4/3}(-2u/(u-1))^{-4/3}  du$$  
 [/INST] $$
du = \frac{2}{(x+1)^2}dx
$$
Also realise that
$$
\left(x^2-1\right)(x+1) = (x-1)(x+1)(x+1) = (x -1)(x+1)^2
$$
$$
\int\frac{1}{\left( (x -1)(x+1)^2\right)^{2/3}}dx = \\
\int\frac{1}{(x+1)^{4/3}(x-1)^{2/3}}dx =\\
\int \frac{(x+1)^{2/3}}{(x+1)^{6/3}}\frac{1}{(x-1)^{2/3}}dx=\\
\int \frac{1}{(x+1)^2}\left(\frac{x+1}{x-1}\right)^{2/3}dx 
$$
Using the sub now 
$$
\frac{1}{2}\int \left(\frac{1}{u}\right)^{2/3}du
$$
Which you can solve.
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