Solve $\int \frac{2x}{(x^2+x+1)^2}dx$ Solve
$$\int \frac{2x}{\left(x^2+x+1\right)^2}dx$$
I tried using integrating in parts by using the partial fraction method where the rational function is split into 
$$\frac{Ax+B}{x^2+x+1} + \frac{Cx+D}{(x^2+x+1)^2}$$
But then I just get the original formula with A=$0$ B=$0$ C=2 and D=$0$...
Please help!

Hint:
$\dfrac{d(x^2+x+1)}{dx}=?$
$$\dfrac{2x}{(x^2+x+1)^2}=\dfrac{2x+1}{(x^2+x+1)^2}-\dfrac1{(x^2+x+1)^2}$$
As $4(x^2+x+1)=(2x+1)^2+3$
how about starting with $2x+1=\sqrt3\tan y$
See also : Trigonometric substitutions, wiki
