# Integral of $\int\frac1{x}\sqrt[3]{\frac{1-x}{1+x}}dx$

I could substitute $t=\sqrt[3]{\frac{1-x}{1+x}}$ and get $\int\frac{6t^3}{t^6-1}dt$, which leads to partial fractions decomposition with 6 variables. That's annoying and may lead to mistakes. Is there any other way to compute this integral?

• I am not sure whether this helps, what if you try $u = t^3$? – flawr Sep 17 '14 at 12:19

The change $t^2=s$ gives $$3\int\frac{s}{s^3-1}\,ds,$$ which is simpler.