integrate $\int\frac{x\cdot dx}{(x^3+1)^2}$ What methods are there to integrate: $$\int\frac{x\cdot dx}{(x^3+1)^2}$$
I know about partial fractions: 
$$\int\frac{x\cdot dx}{(x^3+1)^2} $$
$$= \int\frac{x\cdot dx}{((x+1)(x^2-x+1))^2} $$
$$= \int \left(\frac{A}{x+1}+\frac{Bx+C}{(x+1)^2} + \frac{Dx+E}{x^2-x+1} + \frac{Fx^3+Gx^2+H+I}{(x^2-x+1)^2}\right)dx$$ and after this solving is easy, i was trying to do the same many times, but i can't find coefficients because mistakes or something other.
I want to know about another methods to solve it.
 A: HINT:
First integrate by parts, 
$$3I=\int\dfrac1x\cdot\dfrac{3x^2}{(1+x^3)^2}dx=-\dfrac1{x(1+x^3)}+\int\dfrac{dx}{x^2(1+x^3)}$$
Now use Partial fraction $$\dfrac1{x^2(1+x^3)}=\dfrac Ax+\dfrac B{x^2}+\dfrac C{x+1}+\dfrac{Dx+E}{x^2-x+1}$$
A: $$
\begin{align}
\int\frac{x\,\mathrm{d}x}{\left(x^3+1\right)^2}
&=-\frac13\int\frac1x\,\mathrm{d}\frac1{x^3+1}\tag{1}\\
&=-\frac1{3x\left(x^3+1\right)}-\frac13\int\frac{\mathrm{d}x}{x^2\left(x^3+1\right)}\tag{2}\\
&=-\frac1{3x\left(x^3+1\right)}-\frac13\int\left(\frac1{x^2}-\frac{x}{x^3+1}\right)\,\mathrm{d}x\tag{3}\\
&=-\frac1{3x\left(x^3+1\right)}+\frac1{3x}+\frac19\int\left(\frac{x+1}{x^2-x+1}-\frac1{x+1}\right)\,\mathrm{d}x\tag{4}\\
&=-\frac1{3x\left(x^3+1\right)}+\frac1{3x}-\frac19\log(x+1)+\frac19\int\frac{x-\frac12+\frac32}{\left(x-\frac12\right)^2+\frac34}\,\mathrm{d}x\tag{5}\\
&=\small\frac{x^2}{3\left(x^3+1\right)}-\frac19\log(x+1)+\frac1{18}\log\left(x^2-x+1\right)+\frac1{3\sqrt3}\tan^{-1}\left(\frac{2x-1}{\sqrt3}\right)+C\tag{6}\\
&=\frac{x^2}{3\left(x^3+1\right)}+\frac1{18}\log\left(\frac{x^3+1}{(x+1)^3}\right)+\frac1{3\sqrt3}\tan^{-1}\left(\frac{2x-1}{\sqrt3}\right)+C\tag{7}
\end{align}
$$
Explanation:
$(1)$: prepare to integrate by parts
$(2)$: integrate by parts
$(3)$: partial fractions
$(4)$: partial fractions
$(5)$: integrate
$(6)$: integrate
$(7)$: simplify
