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?
 A: HINT:
$$\int\dfrac{dx}{x^p(1+x^m)^n}=\int\dfrac1{mx^{p+m-1}}\cdot\dfrac{mx^{m-1}}{(1+x^m)^n}dx$$
$$=\dfrac1{mx^{p+m-1}}\int\dfrac{mx^{m-1}}{(1+x^m)^n}dx-\int\left(\dfrac{d\left(\dfrac1{mx^{p+m-1}}\right)}{dx}\cdot\int\dfrac{mx^{m-1}}{(1+x^m)^n}dx\right)dx$$
$$\implies\int\dfrac{dx}{x^p(1+x^m)^n}=-\dfrac1{m(n-1)x^{m+p-1}}\cdot\dfrac1{(1+x^m)^{n-1}}-\dfrac{p+m-1}{m(n-1)}\int\dfrac{dx}{x^{p+m}(1+x^m)^{n-1}}$$
Here start with  $p=0,m=3,n=3$
Then $p=0+3,m=3,n=3-1=2$
Finally use, $$\dfrac1{x^6(1+x^3)}=\dfrac{1-x^6+x^6}{x^6(1+x^3)}=\dfrac{1-x^3}{x^6}+\dfrac1{1+x^3}$$
A: 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 $a$.
A: 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.
