<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 any $A>0$ we have $\int_{-\infty}^{+\infty}\frac{dz}{z^2+A} = \frac{\pi}{\sqrt{A}}$, hence by applying $\frac{d^2}{dA^2}$ to both sides we get
$$ \int_{-\infty}^{+\infty}\frac{dz}{(z^2+A)^3}=\frac{3\pi}{8A^2\sqrt{A}}.\tag{1} $$
On the other hand
$$ \int_{-\infty}^{+\infty}\frac{dx}{(x^2+x+1)^3}\stackrel{x\mapsto z-\frac{1}{2}}{=}\int_{-\infty}^{+\infty}\frac{dz}{\left(z^2+\tfrac{3}{4}\right)^3}=\frac{4\pi}{3\sqrt{3}}.\tag{2}$$
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