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

Question

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

score 4 · Accepted Answer · answered Sep 17 '14 at 12:30

4

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

answered Sep 17 '14 at 12:30

Julián Aguirre

76,354

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

1 Answers1