<s>[INST] Prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$. I have no idea how to do this question.
I'm given $\int^{\infty}_{-\infty}\frac{1}{x^2+2ax+b^2}dx=\frac{\pi}{\sqrt{b^2-a^2}}$ if $b>|a|$ and I'm asked to prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$.
What I've tried:
$\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)}dx=\frac{2\pi}{\sqrt3}$. I've tried setting a variable as the power, ie: $I(r)=\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^r}dx$ but differenciating the inside w.r.t to $r$ doesn't seem to form any differential equation that can be solved.
Answers and hints appreciated!
 [/INST] For real $x$,  we have $x^2 + x + 1 = (x+\frac12)^2 + \frac34 \ge \frac34$. 
This implies for any $t \in (0,\frac34)$, following expansion in $t$ converges:
$$\frac{1}{x^2+x+1 - t} = \sum_{k=0}^\infty \frac{t^k}{(x^2 + x + 1)^{k+1}}$$
Since everything on RHS is non-negative, by Tonelli, we can integrate them term by term:
$$\int_{-\infty}^\infty \frac{dx}{x^2+x+1 - t} = \sum_{k=0}^\infty t^k \int_{-\infty}^\infty \frac{dx}{(x^2+x+1)^{k+1}}
$$
Set $a = \frac12$ and $b = \sqrt{1-t}$. Notice $b > |a|$, the formula you have
tell us the integral on LHS is
$$\frac{\pi}{\sqrt{b^2 - a^2}} = \frac{\pi}{\sqrt{\frac34 - t}}
= \frac{2\pi}{\sqrt{3}\sqrt{1 - \frac{4t}{3}}}$$
Recall $\displaystyle\;\frac{1}{\sqrt{1-4s}}$ is the generating function for the central binomial coefficients: 
$$\frac{1}{\sqrt{1-4s}} = \sum_{k=0}^\infty \binom{2k}{k} s^k$$ 
This leads to
$$\sum_{k=0}^\infty t^k \int_0^\infty \frac{dx}{(x^2+x+1)^{k+1}} =
\frac{2\pi}{\sqrt{3}}\sum_{k=0}^\infty \binom{2k}{k}\frac{t^k}{3^k}
$$
By comparing coefficients of $t^k$ on both sides, we obtain
$$\int_{-\infty}^\infty \frac{dx}{(x^2+x+1)^{k+1}}
= \frac{2\pi}{3^k\sqrt{3}}\binom{2k}{k}\quad\text{ for } k \in \mathbb{N}
$$
In particular, for $k = 2$, this give us
$$\int_{-\infty}^\infty \frac{dx}{(x^2+x+1)^3} = \frac{2\pi}{3^2\sqrt{3}}\binom{4}{2} = \frac{4\pi}{3\sqrt{3}}$$
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