<s>[INST] Computing $\int \frac{1}{(1+x^3)^3}dx$ I tried various methods like trying to break it into partial fractions after factorizing, applying substitutions but couldn't think of any. How will we integrate this?
 [/INST] Hint. By writing
$$
1+x^3=(1+x)(1-x+x^2), \qquad (1+x^3)^3=(1+x)^3(1-x+x^2)^3,
$$ one may observe that there is a partial fraction decomposition of the form
$$
\begin{align}
\frac{1}{(1+x^3)^3}&=\frac{a_1}{(1+x)}+\frac{a_2}{(1+x)^2}+\frac{a_3}{(1+x)^3}
\\\\&+\frac{\alpha_1x+\beta_1}{1-x+x^2}+\frac{\alpha_2x+\beta_2}{(1-x+x^2)^2}+\frac{\alpha_3x+\beta_3}{(1-x+x^2)^3}
\end{align}
$$ then each term can be classically integrated. One may notice that
$$
\frac{\alpha_3x+\beta_3}{(1-x+x^2)^3}=\frac12\frac{\alpha_3(2x-1)+2\beta_3}{(1-x+x^2)^3}+\frac12\frac{\alpha_3+2\beta_3}{(1-x+x^2)^3}
$$ which is easier to integrate and that
$$
(1-x+x^2)^3=((x-1/2)^2+3/4)^3=(3/4)^3(u^2+1)^3,
$$ the changes of variable $ x-1/2=\sqrt{3}u/2$, $u=\tan t$, will turn out useful.
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