$\int_{-\infty}^\infty\frac{1+x^2}{1+x^4}\,dx$

The integrand is even, so \[ I=2\int_0^\infty\frac{1+x^2}{1+x^4}\,dx. \] Under the substitution \(x=1/t\), \[ \int_0^\infty\frac{x^2}{1+x^4}\,dx =\int_0^\infty\frac{1}{1+t^4}\,dt. \] Hence \[ I=4\int_0^\infty\frac{dx}{1+x^4}. \] The standard residue computation in the upper half-plane gives \[ \int_{-\infty}^{\infty}\frac{dx}{1+x^4} =2\pi i\left( \operatorname{Res}_{z=e^{i\pi/4}}\frac1{1+z^4} +\operatorname{Res}_{z=e^{3i\pi/4}}\frac1{1+z^4} \right) =\frac{\pi}{\sqrt2}. \] Therefore \[ \int_0^\infty\frac{dx}{1+x^4}=\frac{\pi}{2\sqrt2}, \] and so \[ \boxed{I=\sqrt2\,\pi.} \]