Integrate $\int\frac{1}{x^3+1}dx$ The problem is, as stated:
$$\int\frac{1}{x^3+1}dx$$
I tried using substitution: $t^3 = x^3 + 1$ but didn't get far with that. I also tried setting: $t = x^3 + 1$, with no luck again.
I tried partial decomposition but I didn't know how to integrate $$\int\frac{1}{x^2-x+1}$$ and I kept getting that term when expanding $x^3 + 1$
Any help would be much appreciated.

$$\int\frac {1}{x^2-x+1} dx=\int\frac {1}{(x-\frac{1}{2})^2 +\frac{3}{4}}dx$$
And that take $x-\frac{1}{2}=t$.
Always use $ax^2+bx+c=a(x+\frac{b}{2a}) ^2 - \frac{b^2-4ac}{4a}$
