<s>[INST] How to Prove $\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$ Question:-  Prove that
$\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$
On factoring the denominator we get,
$\int_{0}^{\infty}\frac {1}{(x^4+x^2+1)(x^4-x^2+1)}dx$
Partial fraction of the integrand contains big terms with their long integral.So i didn't proceed with partial fraction.I'm unable figure out any other method.I think that there might be some other method for evaluation of this definite integral since its value is
$\frac{π}{2\sqrt{3}}$.
Does anyone have nice way to solve it!
 [/INST] Taking $x\mapsto \frac{1}{x}$ transforms the integral into $I= \int_0^{\infty} \frac{x^6}{1+x^4+x^8} d x$ and then taking the average of them gives $$
\begin{aligned}
I & =\frac{1}{2} \int_0^{\infty} \frac{x^6+1}{x^8+x^4+1} d x \\
& =\frac{1}{2} \int_0^{\infty} \frac{\left(x^2+1\right)\left(x^4-x+1\right)}{\left(x^4+x^2+1\right)\left(x^4-x+1\right)} d x \\
& =\frac{1}{2} \int_0^{\infty} \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1} d x \\
& =\frac{1}{2} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+3} \\
& =\frac{1}{2 \sqrt{3}}\left[\tan ^{-1}\left(\frac{\left.x-\frac{1}{x}\right)}{\sqrt{3}}\right)\right]_0^{\infty} \\
& =\frac{\pi}{2 \sqrt{3}}
\end{aligned}
$$
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