3
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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?

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  • $\begingroup$ I am not sure whether this helps, what if you try $u = t^3$? $\endgroup$ – flawr Sep 17 '14 at 12:19
4
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The change $t^2=s$ gives $$ 3\int\frac{s}{s^3-1}\,ds, $$ which is simpler.

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