Computing the integral of a rational function I need to compute the integral $\int \dfrac{2x}{(x^2+x+1)^2} \cdot dx$.  I tried using the integration of a rational function technique, with $\frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}$, but this simply returned $C=2$ and $A,B,D = 0$, so it doesn't really change anything.  
I also tried using a $u$ substitution, setting $u=x^2+x+1$.  This made the numerator $2x=\frac{du}{dx}-1$, but I'm not really sure if I can do that/how to solve an integral with a derivative as a part of it.
How would I go about solving this?
Thanks for your time.
 A: $$2\int { \frac { x }{ (x^ 2+x+1)^ 2 }  } $$
$$2\int { \frac { x }{ ((x+\frac { 1 }{ 2 } )^ 2+\frac { 3 }{ 4 } )^ 2 }  } $$
Apply u-substitution: $u=x+\frac12$
$$2\int { \frac { 8(2u-1) }{ (4u^ 2+3)^ 2 }  } du$$
$$2(8)\int { \frac { 2u-1 }{ (4u^ 2+3)^ 2 }  } du$$
Apply the Sum Rule
$$2(8)(\int { \frac { 2u }{ (4u^ 2+3)^ 2 }  } du-\int { \frac { 1 }{ (4u^ 2+3)^ 2 } du } )$$
Now, $$\int { \frac { 2u }{ (4u^ 2+3)^ 2 }  } =- { \frac { 1 }{ 4(4u^ 2+3) }  } $$
Now, $$\int { \frac { 1 }{ (4u^ 2+3)^ 2 } du=\frac { 1 }{ 12\sqrt { 3 }  } (arctan(\frac { 2 }{ \sqrt { 3 }  } u)+\frac { 1 }{ 2 } sin(2arctan(\frac { 2 }{ \sqrt { 3 }  } u))) } $$
$$=2(8)(- { \frac { 1 }{ 4(4u^ 2+3) }  }- { \frac { 1 }{ 12\sqrt { 3 }  } (arctan(\frac { 2 }{ \sqrt { 3 }  } u)+\frac { 1 }{ 2 } sin(2arctan(\frac { 2 }{ \sqrt { 3 }  } u))) } $$
After doing small calculations and substituting $u=x+\frac12$,
$$\int\frac{2x}{x^2+x+1}dx=16(-\frac{1}{4(4x^2+4x+4)}-\frac{1}{24\sqrt{3}}(2arctan(\frac{2x+1}{\sqrt{3}})+sin(2arctan(\frac{2x+1}{\sqrt{3}}))))+C$$
A: Apply integration by parts to 
$$\int\frac{1}{x^2+x+1} dx$$ and extract your desired integral from the result. 
