How to find $$\int\frac{{\operatorname d}x}{1-x^3}?$$
Is it possible by sine or cosine functions?
It is not easy to calculate it by reparametrization.
2026-04-30 04:09:25.1777522165
Evaluating $\int\frac{ {\operatorname d}x}{1-x^3}$
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Note that $$1-x^3=(1-x)(1+x+x^2).$$
Then, you'll find $$\frac{1}{1-x^3}=\frac{1/3}{1-x}+\frac{(1/3)x+(2/3)}{x^2+x+1}.$$