Evaluation of $\int \:\frac{1}{\sqrt[3]{\left(1+x\right)^2\left(1-x\right)^4}}dx$ using trig substitution

Question

Recently I came accross this integral :
$$ \int \:\frac{1}{\sqrt[3]{\left(1+x\right)^2\left(1-x\right)^4}}dx $$
I would evaluate it like this, first start by the substitution:
$$ x=\cos(2u) $$
$$ dx=-2\sin(2u)du $$
Our integral now becomes:
$$\int \:\frac{-2\sin \left(2u\right)du}{\sqrt[3]{\left(\cos \left(2u\right)+1\right)^2\left(\cos \:\left(2u\right)-1\right)^4}}$$
$$\cos(2u)=\cos(u)^2-\sin(u)^2$$
Thus:
$$\cos(2u)+1=2\cos(u)^2$$
$$\cos(2u)-1=-2\sin(u)^2$$
Thus our integral now becomes:
$$\int \:\frac{-\sin \left(2u\right)du}{\sqrt[3]{4\cos \left(u\right)^416\sin \left(u\right)^8}}=\frac{1}{2}\int \:\frac{-\sin \left(2u\right)du}{\sqrt[3]{\cos \left(u\right)^4\sin \left(u\right)^8}}$$
we know:
$$\sin \left(u\right)=\cos \left(u\right)\tan \left(u\right)$$
Thus our integral becomes:
$$\int \:\frac{-\tan \left(u\right)\cos \left(u\right)^2du}{\cos \:\left(u\right)^4\sqrt[3]{\tan \left(u\right)^8}}=\int \frac{-\tan \:\left(u\right)\sec \left(u\right)^2du}{\sqrt[3]{\tan \:\left(u\right)^8}}\:$$
By letting $$v=\tan \:\left(u\right)$$
$$dv=\sec \left(u\right)^2du$$
Our integral now becomes:
$$\int -v\:^{1-\frac{8}{3}}dv=-\frac{v^{2-\frac{8}{3}}}{2-\frac{8}{3}}+C=\frac{3}{2\sqrt[3]{v^2}}+C$$
Undoing all our substitutions:
$$\frac{3}{2\sqrt[3]{\tan \left(u\right)^2}}+C$$
$$\tan \:\left(u\right)^2=\frac{1}{\cos \left(u\right)^2}-1=\frac{2}{1+\cos \left(2u\right)}-1=\frac{2}{1+x}-1$$
Our integral therefore:
$$\frac{3}{2\sqrt[3]{\frac{2}{1+x}-1}}+C$$
However, online integral give me anti-derivative of $$\frac{-3\sqrt[3]{\frac{2}{x-1}+1}}{2}+C$$ so I want to know where I went wrong

Answer 1

Your result is correct, which can be obtained alternatively by substituting $t= \frac{1+x}{1-x}$ to arrive at

$$ \int \:\frac{dx}{\sqrt[3]{\left(1+x\right)^2\left(1-x\right)^4}} =\frac12\int t^{-2/3}dt = \frac32 t^{1/3}+C=\frac32 \sqrt[3]\frac{1+x}{1-x} +C$$