<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] Let $(a,b,c,d)$ to be the complex roots of $x^8+x^4+1=0$. So, after partial fraction decomposition, the integrand write
$$\frac{1}{(a-b) (a-c) (a-d) \left(x^2-a\right)}+\frac{1}{(b-a) (b-c) (b-d) \left(x^2-b\right)}+$$
$$\frac{1}{(c-a) (c-b) (c-d) \left(x^2-c\right)}+\frac{1}{(d-a) (d-b) (d-c) \left(x^2-d\right)}$$
This makes the definite integral to be
$$\frac \pi 2\frac{   \left(\sqrt{a} \left(\sqrt{b}+\sqrt{c}+\sqrt{d}\right)^2+a
   \left(\sqrt{b}+\sqrt{c}+\sqrt{d}\right)+\left(\sqrt{b}+\sqrt{c}\right)
   \left(\sqrt{b}+\sqrt{d}\right) \left(\sqrt{c}+\sqrt{d}\right)\right)}{ \sqrt{a}
   \sqrt{b} \sqrt{c} \sqrt{d} \left(\sqrt{a}+\sqrt{b}\right)
   \left(\sqrt{a}+\sqrt{c}\right) \left(\sqrt{a}+\sqrt{d}\right)
   \left(\sqrt{b}+\sqrt{c}\right) \left(\sqrt{b}+\sqrt{d}\right)
   \left(\sqrt{c}+\sqrt{d}\right)}$$ Now, using
$$a=\frac{1+i \sqrt{3}}{2}\qquad b=\frac{1-i \sqrt{3}}{2}\qquad c=-\frac{1+i \sqrt{3}}{2}\qquad d=-\frac{1-i \sqrt{3}}{2}$$ leads to the result.
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