My attempt:
I am considering a contour C consisting of upper half circle $|z|=R$ and the real axis from $-R$ to $R$ and finding $\int_C \dfrac{1}{z^6+1}dz$ around it.
First, I find the poles of this function, which are given by $z^6+1=0$
The poles are all simple poles, $z=\pm i,\dfrac{\sqrt{3}}{2} \pm i\dfrac{1}{2},-\dfrac{\sqrt{3}}{2} \pm i\dfrac{1}{2} $
Only the 3 poles $z= i,\dfrac{\sqrt{3}}{2} + i\dfrac{1}{2},-\dfrac{\sqrt{3}}{2} + i\dfrac{1}{2} $ lie within the contour, so I find the residues at these poles and then apply residue theorem to find $\int_C \dfrac{1}{z^6+1}dz$
Residue at $z=i$ comes out to be $\dfrac{1}{6i}$, at $z=\dfrac{\sqrt{3}}{2} + i\dfrac{1}{2}$ comes out to be $\dfrac{1}{3i(1-\sqrt3 i)}$ and at $z=\dfrac{\sqrt{3}}{2} + i\dfrac{1}{2}$ is $\dfrac{-1}{6i}$
So from Cauchy's theorem I get $\int_C \dfrac{1}{z^6+1}dz=2\pi i \times $sum of the residues= $\dfrac{2\pi}{3(1-\sqrt3 i)}$
Taking $R \to \infty$ and using $f(z)$ is an even function, I get my final answer as: $\int_0^{\infty} \frac{1}{1+x^6} dx=\dfrac{\pi}{3(1-\sqrt3 i)}$
Is this solution correct? Is there a shorter or better way to do this? Thank you.