Calculating $\int \frac{dx}{\sqrt[3]{(x+1)^2(x-1)^7}}$ I am trying to calculate the integral
$$
\int \frac{dx}{\sqrt[3]{(x+1)^2(x-1)^7}}.
$$
I know that the answer is
$$
\frac3{16}(3x-5)\sqrt[3]{\frac{x+1}{(x-1)^4}}+C;
$$
this led me to the idea of introducing the new variable
$$
t^3=\frac{x+1}{(x-1)^4}.
$$
But that got me nowhere, because I can't express $x$ in terms of $t$.

Substitute $t^3=\frac{x+1}{x-1}$. Then, $x=\frac{t^3+1}{t^3-1}$, $dx= -\frac{6t^2}{(t^3-1)^2}dt$ and
\begin{align}
&\int \frac{1}{\sqrt[3]{(x+1)^2(x-1)^7}}dx\\
=&-\frac34 \int (t^3-1)dt = -\frac3{16}t^4 +\frac34t +C\\
= & -\frac3{16}\left(\frac{x+1}{x-1} \right)^{\frac43} +\frac34 \left(\frac{x+1}{x-1} \right)^{\frac13} +C\\
=& \frac3{16}(3x-5)\sqrt[3]{\frac{x+1}{(x-1)^4}}+C
\end{align}
