<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. Another path. Consider a fixed real number $a>0$. From
$$
a^3+x^3=(a+x)(a^2-ax+x^2)
$$ one gets
$$
\begin{align}
\frac{1}{a^3+x^3}&=\frac{1}{3 a^2 (a+x)}+\frac{2 a-x}{3 a^2 \left(a^2-a x+x^2\right)}
\end{align}
$$ integrate the preceding identity to get
$$
\int\frac{dx}{a^3+x^3}=\frac1{3a^2}\ln(a+x)-\frac1{6a^2}\ln(a^2-a x+x^2)+\frac1{a^2\sqrt{3} }\arctan\left(\frac{2x-a}{a\sqrt{3}}\right)
$$ put $a^3 \to a$ and just differentiate twice with respect to $a$.
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