Integrate $\int ((x^2-1)(x+1))^{-2/3} \, dx$ using $u = \tan^{-1}(x)$ The problem asks to solve $\int ((x^2-1)(x+1))^{-2/3} \, dx$ using the u-substitution $u = \tan^{-1}(x)$.
I was able to solve the integral using the $u$-substitution $v = \frac{x-1}{x+1}$ (see below); however, I was not able to make any progress when $u = \tan^{-1}(x)$ except for finding that $du = \frac{1}{1+x^2} dx$. Any hints would be greatly appreciated.
$$\int ((x^2-1)(x+1))^{-2/3} \, dx = \int \frac{1}{\Big(\frac{x-1}{x+1}\Big)^{2/3}(x+1)^2} \, dx = \frac{1}{2}\int {v}^{-2/3} \, dv = \frac{3}{2}\Big(\frac{x-1}{x+1}\Big)^{1/3}+c$$

I think I have figured out a messy way to get the solution. Let $u = \tan^{-1}(x)$.
$\begin{align*}
\int \frac{1}{(x^2-1)^{2/3}(x+1)^{2/3}} \, dx &= \int \frac{1+\tan^2(u)}{(\tan^2(u)-1)^{2/3}(\tan(u)+1)^{2/3}} \, du \\
&= \int \frac{\sec^2(u)}{(\tan^2(u)-1)^{2/3}(\tan(u+\frac{\pi}{4})(1-\tan(u)))^{2/3}} \, du\\
&= \int \frac{1}{(\tan(u+\frac{\pi}{4}))^{2/3}(\cos(u)-\sin(u))^2\Big(\frac{\cos(u)+ \sin(u)}{\cos(u)-\sin(u)}\Big)^{2/3}} \, du\\
&= \int \frac{1}{(\cos(u)-\sin(u))^2(\tan(u+\frac{\pi}{4}))^{4/3}} \, du\\
&= \int \frac{\sec^2(u+\frac{\pi}{4})}{2(\tan(u+\frac{\pi}{4}))^{4/3}} \, du\\
&= -\frac{3}{2}(\tan(u+\frac{\pi}{4}))^{-1/3} + c \\
&= \frac{3}{2}\Big(\frac{x-1}{x+1}\Big)^{1/3}+c\\
\end{align*}$
Please let me know if you find an easier solution!
