Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$ Let

$$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$
  $$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$

Prove that $I=J={\pi \over 2\sqrt3}$

Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$
$x=\infty \rightarrow u={\pi\over 2}$, $x=0\rightarrow u=0$
Rewrite $(1)$ as
$$I=\int_{0}^{\infty}{1\over (1+x^2)^2-x^2}dx$$
then
$$\int_{0}^{\pi/2}{\sec^2{u}\over \sec^4{u}-\tan^2{u}}du\tag3$$
Simplified to
$$I=\int_{0}^{\pi/2}{1\over \sec^2{u}-\sin^2{u}}du\tag4$$
Then to
$$I=2\int_{0}^{\pi/2}{1+\cos{2u}\over (2+\sin{2u})(2-\sin{2u})}du\tag5$$
Any hints on what to do next?

Re-edit (Hint from Marco)
$${1\over x^8+x^4+1}={1\over 2}\left({x^2+1\over x^4+x^2+1}-{x^2-1\over x^4-x^2+1}\right)$$
$$M=\int_{0}^{\infty}{x^2+1\over x^4+x^2+1}dx=\int_{0}^{\infty}{x^2\over x^4+x^2+1}dx+\int_{0}^{\infty}{1\over x^4+x^2+1}dx={\pi\over \sqrt3}$$
$$N=\int_{0}^{\infty}{x^2-1\over x^4-x^2+1}dx=0$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx={1\over 2}\left({\pi\over \sqrt3}-0\right)={\pi\over 2\sqrt3}.$$

Both integrands are even, so the integrals can be extended to negative infinity. The integrands decay quickly enough for the contour to be closed with a semicircle at infinity in the upper half plane without chaning the integral. Then the integrals can be determined with the residue theorem.
Multiplying the denominator of the first integrand by $x^2-1$ yields $x^6-1$, so there are simple poles at $x=\omega_k=\exp\left(\frac{2\pi\mathrm i k}6\right)$ for $k=1,2,4,5$. The ones for $k=1,2$ lie in the upper half plane, and the corresponding residues are
$$
r_k=\lim_{x\to\omega_k}\frac{\left(x^2-1\right)\left(x-\omega_k\right)}{x^6-1}=\lim_{x\to\omega_k}\frac{(x-\omega_0)(x-\omega_3)(x-\omega_k)}{\prod_{j=0}^5(x-\omega_j)}\;,
$$
with
$$
r_1=\frac1{(\omega_1-\omega_2)(\omega_1-\omega_4)(\omega_1-\omega_5)}=-\frac14-\frac1{4\sqrt3}\mathrm i
$$
and
$$
r_2=\frac1{(\omega_2-\omega_1)(\omega_2-\omega_4)(\omega_2-\omega_5)}=\frac14-\frac1{4\sqrt3}\mathrm i
$$
Thus the contour integral is
$$
2\pi\mathrm i\left(\left(-\frac14-\frac1{4\sqrt3}\mathrm i\right)+\left(\frac14-\frac1{4\sqrt3}\mathrm i\right)\right)=\frac\pi{\sqrt3}\;,
$$
and your real integral is half of that.
For the second integral, multiplying the denominator by $x^4-1$ yields $x^{12}-1$, so there are simple poles at $\nu_k=\exp\left(\frac{2\pi\mathrm ik}{12}\right)$ for $k=1,2,4,5,7,8,10,11$, of which the ones for $k=1,2,4,5$ lie in the upper half plane. I'll leave it to you to find and add their residues as above.
