How to evaluate $\int\sqrt[3] {\frac{1}{(x+1)^2(x-1)^4}} dx$? My integral is
$$I=\int\sqrt[3] {\frac{1}{(x+1)^2(x-1)^4}} dx$$
and hence
$$I=\int\frac{1}{(x-1)(x+1)}\sqrt[3] {\frac{x+1}{x-1}}dx $$
$\cos2\theta$ substitution wont be helpful here because of the cube root. Should I apply by parts ?

First notice that
$$\eqalign{
   I &= \int {\root 3 \of {{1 \over {{{(x + 1)}^2}{{(x - 1)}^4}}}} } dx  \cr 
  &= \int {{1 \over {\left( {x - 1} \right)\left( {x + 1} \right)}}\root 3 \of {{{x + 1} \over {x - 1}}} } dx  \cr 
  &= \int {{1 \over {{{\left( {x - 1} \right)}^2}}}{{x - 1} \over {x + 1}}\root 3 \of {{{x + 1} \over {x - 1}}} } dx \cr} $$
Now, make the substitution
$$\eqalign{
   u &= {{x + 1} \over {x - 1}}  \cr 
   du &=  - {2 \over {{{\left( {x - 1} \right)}^2}}}dx \cr} $$
to obtain
$$\eqalign{
I &=  - {1 \over 2}\int {{1 \over u}\root 3 \of u du}   \cr 
  &=  - {1 \over 2}\int {{u^{ - {2 \over 3}}}du}  \cr
  &=  - {3 \over 2}{u^{{1 \over 3}}} + C \cr
  &=  \boxed{ - {3 \over 2}{\left( {{{x + 1} \over {x - 1}}} \right)^{{1 \over 3}}} + C} \cr} $$
