$\int_{\RR}\frac{dx}{1+x^4}$ and the poles of $\frac{1}{1+z^4}$

The poles are the four roots of \[ z^4=-1=e^{i(\pi+2\pi k)}, \] namely \[ \boxed{ e^{i\pi/4},\quad e^{3i\pi/4},\quad e^{5i\pi/4},\quad e^{7i\pi/4}.} \] All are simple.

To evaluate the integral, integrate \[ F(z)=\frac1{1+z^4} \] over the upper semicircle of radius \(R\) and let \(R\to\infty\). The arc integral vanishes because \(F(z)=O(R^{-4})\). The enclosed poles are \[ \zeta_1=e^{i\pi/4}, \qquad \zeta_2=e^{3i\pi/4}. \] Since the poles are simple, \[ \operatorname{Res}(F;\zeta)=\frac1{4\zeta^3}. \] Hence \[ \operatorname{Res}(F;\zeta_1) +\operatorname{Res}(F;\zeta_2) =\frac1{4}\left(e^{-3i\pi/4}+e^{-9i\pi/4}\right) =-\frac{i}{2\sqrt2}. \] Therefore \[ \int_{-\infty}^{\infty}\frac{dx}{1+x^4} =2\pi i\left(-\frac{i}{2\sqrt2}\right) =\boxed{\frac{\pi}{\sqrt2}}. \]