Integrate $\int\frac{x^3}{(x^3+1)^2}dx$ any ideas on how I could continue this integral
$$\int\frac{x^3}{(x^3+1)^2}dx$$
I am half way done, by entirely using fraction decomposition
$\int\frac{x^3}{(x^3+1)^2}dx=\int\frac{1}{x^3+1}dx-\int\frac{1}{(x^3+1)^2}dx$
$\frac{1}{x^3+1}=\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}$ from which it follows that $A=\frac{1}{3}, B=-\frac{1}{3}, C=\frac{2}{3}$
$\frac{1}{x^3+1}=\frac{1}{3(x+1)}-\frac{x-2}{3(x^2-x+1)}$
$\int\frac{1}{3(x+1)}dx=\frac{1}{3}\ln|{x+1}|+C$
$\int\frac{x-2}{3(x^2-x+1)}dx=\frac{1}{3}\left(\int\frac{2x-1}{2(x^2-x+1)}dx-\int\frac{3}{2(x^2-x+1)}dx\right)=\frac{1}{3}\left(\frac{1}{2}\ln|{x^2-1+1}|-\frac{3}{2}\frac{2}{\sqrt{3}}\arctan(\frac{2x-1}{\sqrt{3}})\right)+C$
$\int\frac{dx}{(x^3+1)^2}=?$
This is the part where I am stuck currently, using fraction decomposition seems to take a lot of time and effort, could I be missing anything?

The polynomial $x^3+1$ is not irreducible (as you have seen),
$$x^3+1=(x+1)(x^2-x+1).$$
You need to continue decomposing in order to compute the integral.
