Calculating a real integral using complex integration $$\int^\infty_0 \frac{dx}{x^6 + 1}$$
Does someone know how to calculate this integral using complex integrals? I don't know how to deal with the $x^6$ in the denominator.

Thankfully the integrand is even, so we have
$$
\int^\infty_0 \frac{dx}{x^6 + 1} = \frac{1}{2}\int^\infty_{-\infty} \frac{dx}{x^6 + 1}.
\tag{1}
$$
To find this, we will calculate the integral
$$
\int_{\Gamma_R} \frac{dz}{z^6+1},
$$
where $\Gamma_R$ is the semicircle of radius $R$ in the upper half-plane, $C_R$, together with the line segment between $z=-R$ and $z=R$ on the real axis.

(Image courtesy of Paul Scott.)
Then
$$
\int_{\Gamma_R} \frac{dz}{z^6+1} = \int_{-R}^{R} \frac{dx}{x^6+1} + \int_{C_R} \frac{dz}{z^6+1}.
$$
We need to show that the integral over $C_R$ vanishes as $R \to \infty$.  Indeed, the triangle inequality gives
$$\begin{align}
\left| \int_{C_R} \frac{dz}{z^6+1} \right| &\leq L(C_R) \cdot \max_{C_R} \left| \frac{1}{z^6+1} \right| \\
           &\leq \frac{\pi R}{R^6 - 1},
\end{align}$$
where $L(C_R)$ is the length of $C_R$.  From this we may conclude that
$$
\lim_{R \to \infty} \int_{\Gamma_R} \frac{dz}{z^6+1} = \int_{-\infty}^{\infty} \frac{dx}{x^6+1}.
\tag{2}
$$
The integral on the left is evaluated by the residue theorem.  For $R > 1$ we have
$$
\int_{\Gamma_R} \frac{dz}{z^6+1} = 2\pi i \sum_{k=0}^{2} \operatorname{Res}\left(\frac{1}{z^6+1},\zeta^k \omega\right),
$$
where $\zeta$ is the primitive sixth root of unity and $\omega = e^{i\pi/6}$.  Note that this is because $\omega$, $\zeta\omega$, and $\zeta^2 \omega$ are the only poles of the integrand inside $\Gamma_R$.  The sum of the residues can be calculated directly, and we find that
$$
\int_{\Gamma_R} \frac{dz}{z^6+1} = 2\pi i \sum_{k=0}^{2} \operatorname{Res}\left(\frac{1}{z^6+1},\zeta^k \omega\right) = \frac{\pi}{3 \sin(\pi/6)} = \frac{2\pi}{3}.
$$
Thus, from $(1)$ and $(2)$ we conclude that
$$
\int_{0}^{\infty} \frac{dx}{x^6+1} = \frac{\pi}{3}.
$$
In general,
$$
\int_{0}^{\infty} \frac{dx}{x^{2n}+1} = \frac{\pi}{2 n \sin\left(\frac{\pi}{2n}\right)}
$$
for $n \geq 1$.
