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!

Since your function is even, your integral is$$\require{cancel}\frac12\int_{-\infty}^\infty\frac{\mathrm dx}{x^8+x^4+1}.$$On the other hand, $x^8+x^4+1=\dfrac{x^{12}-1}{x^4-1}$ and therefore the roots of $x^8+x^4+1$ are the roots of order $12$ of $1$ which are not fourth roots of $1$. Among these, those with imaginary part greater than $0$ are $e^{\pi i/6}$, $e^{\pi i/3}$, $e^{2\pi i/3}$ and $e^{5\pi i/6}$. The residue of $\dfrac1{z^8+z^4+1}$ at these points is, respectively, $-\dfrac1{4\sqrt3}$, $-\dfrac i{4\sqrt3}$, $-\dfrac i{4\sqrt3}$, and $\dfrac1{4\sqrt3}$. Therefore, if $R>1$, then, by the residue theorem\begin{multline}\int_{-R}^R\frac{\mathrm dx}{x^8+x^4+1}+\int_{|z|=R,\ \operatorname{Im}z\geqslant0}\frac{\mathrm dz}{z^8+z^4+1}=\\=2\pi i\left(\cancel{-\frac1{4\sqrt3}}-\frac i{4\sqrt3}-\frac i{4\sqrt3}+\cancel{\frac1{4\sqrt3}}\right)=\frac\pi{\sqrt3}\end{multline}and so, since$$\lim_{R\to\infty}\int_{|z|=R,\ \operatorname{Im}z\geqslant0}\frac{\mathrm dz}{z^8+z^4+1}=0,$$we have\begin{align}\int_{-\infty}^\infty\frac{\mathrm dx}{x^8+x^4+1}&=\lim_{R\to\infty}\int_{-R}^R\frac{\mathrm dx}{x^8+x^4+1}\\&=\frac\pi{\sqrt3}.\end{align}
