$\int_{-\infty}^\infty\frac{dx}{1+x^4}=\frac{\pi}{\sqrt2}$

The poles of \[ \frac1{1+z^4} \] are the four roots of \(z^4=-1=e^{i\pi}\), namely \[ e^{i\pi/4},\quad e^{3i\pi/4},\quad e^{5i\pi/4},\quad e^{7i\pi/4}. \] Integrate over the upper semicircle. The arc contribution tends to \(0\) because the integrand is \(O(R^{-4})\) while the arc length is \(O(R)\). The poles in the upper half-plane are \[ z_1=e^{i\pi/4},\qquad z_2=e^{3i\pi/4}. \] They are simple, and \[ \operatorname{Res}_{z=z_j}\frac1{1+z^4}=\frac1{4z_j^3}. \] Hence \[ \int_{-\infty}^{\infty}\frac{dx}{1+x^4} =2\pi i\left(\frac1{4z_1^3}+\frac1{4z_2^3}\right). \] Since \[ z_1^{-3}=e^{-3i\pi/4}=\frac{-1-i}{\sqrt2}, \qquad z_2^{-3}=e^{-9i\pi/4}=\frac{1-i}{\sqrt2}, \] the sum is \(-\sqrt2\,i\). Therefore \[ \int_{-\infty}^{\infty}\frac{dx}{1+x^4} =2\pi i\left(-\frac{\sqrt2\,i}{4}\right) =\frac{\pi}{\sqrt2}. \]