How to evaluate $\int\sqrt[3] {\frac{1}{(x+1)^2(x-1)^4}} dx$?

Question

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 ?

Answer 1

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

$$\eqalign{ I &= - {1 \over 2}\int {{1 \over u}\root 3 \of u du} \cr &= - {1 \over 2}\int {{u^{ - {2 \over 3}}}du} \cr &= - {3 \over 2}{u^{{1 \over 3}}} + C \cr &= \boxed{ - {3 \over 2}{\left( {{{x + 1} \over {x - 1}}} \right)^{{1 \over 3}}} + C} \cr} $$