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!

Use $x^8+x^4+1=(x^4+x^2+1)(x^4-x^2+1)$ to partial decompose
\begin{align} \int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx
&=\frac12\int_{0}^{\infty}\frac {x^2+1}{x^4+x^2+1}dx
 - \frac12\int_{0}^{\infty}\frac {x^2-1}{x^4-x^2+1}dx\\
&= \frac12\int_{0}^{\infty}\frac {d(1-\frac1x)}{(x-\frac1x)^2+3}
 - \frac12\int_{0}^{\infty}\frac {d(1+\frac1x)}{(x+\frac1x)^2-3}\\
&=\frac12\cdot \frac{π}{\sqrt{3}}+\frac12\cdot0= \frac{π}{2\sqrt{3}}
\end{align}
