Solve $\int \frac{2x}{(x^2+x+1)^2}dx$

Question

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!

Answer 1

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