Give me a hint to solve this integral $$\int_0^1 \frac{1}{(x^2+x+1)^2} dx$$
I don't have any clue as to where to even start. I am struggling on this one. 
Please shed some light by giving me a Hint
 A: Hint: Since $\frac{1}{(1-z)^2}=\sum_{n\geq 1}n z^{n-1}$,
$$\begin{eqnarray*}\int_{0}^{1}\frac{(1-x)^2}{(1-x^3)^2}\,dx &=& \sum_{n\geq 1}\int_{0}^{1}n(1-x)^2 x^{3n-3}\,dx = \sum_{n\geq 1}\int_{0}^{1}n\left(x^{3n-3}-2x^{3n-2}+x^{3n-1}\right)\\&=&\sum_{n\geq 1}n\left(\frac{1}{3n-2}-\frac{2}{3n-1}+\frac{1}{3n}\right)=\frac{2}{3} \sum_{n\geq 0}\frac{1}{(3n+1)(3n+2)}.\end{eqnarray*}$$
In order to compute the last series, you may exploit the following identities involving the digamma function:
$$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b},\qquad \psi(1-z)-\psi(z)=\pi\cot(\pi z).$$
Obviously, this is not the only approach. You may also notice that the last series is related with a Dirichlet $L$-function evaluated at $s=1$; such a series is $L_{\chi}(1)$ where $\chi$ is the Legendre symbol $\pmod{3}$, and we may also exploit the class number formula $(2)+(5)$ in the link.
A: $$\dfrac{d\left[\dfrac{ax^2+bx+c}{x^2+x+1}\right]}{dx}$$
$$=\dfrac{(x^2+x+1)(2ax+b)-(ax^2+bx+c)(2x+1)}{(x^2+x+1)^2}$$
We need $1=(x^2+x+1)(2ax+b)-(ax^2+bx+c)(2x+1)=(a-b)x^2+(a-2c)x+b-c$
$\implies a-b=0,a-2c=0,b-c=1\implies \cdots$
A: HINT:
Use Trigonometric substitution
$$x^2+x+1=\dfrac{(2x+1)^2+(\sqrt3)^2}4$$
Set $2x+1=\sqrt3\tan\theta$
