<s>[INST] How to integrate $\int_0^\infty \frac{1}{1+y^4} dy$ I tried the trigonometric substitution $y^2 = \tan \theta, sec^2\theta = 1 + y^4$
But now I'm stuck with $\frac12 \int \frac{\sqrt{\sin \theta}}{(\cos\theta)^{\frac92} } d \theta$
I ran out of imagination as what to try now
 [/INST] $$\int_0^\infty \frac{dx}{x^4+1} = \frac{1}{2}\int_{-\infty}^\infty \frac{dx}{x^4+1}$$
The latter integral is trivial by means of contour integration.
Let $C$ be the canonical semicircle contour, along the real axis from $-R$ to $-R$ and around the semicircle $Re^{i\theta}$ for $\theta \in [0,\pi]$.  Letting $R \to \infty$, we consider the function on the arc ($|z| = R$):
$$\left|\frac{1}{x^4+1}\right|\le  \frac{1}{|x^4+1|} \le \frac{1}{|x^4|-1} \le \frac{2}{R^4} \to 0$$
so by the estimation lemma, the integral around the arc disappears.
$$\oint_C \frac{dz}{z^4+1} = \int_{-\infty}^\infty \frac{dx}{x^4+1} =2 \pi i \sum \operatorname*{Res}\frac{1}{z^4+1}$$
where the residues are of poles in the upper half plane.  These poles are $z_1=e^{i\pi/4}$ and $z_2=e^{3i\pi/4}$.  It follows that
$$b_1=\operatorname*{Res}_{z=z_1}\frac{1}{z^4+1} = -\frac{1}{4} e^{i\pi/4}$$
$$b_2=\operatorname*{Res}_{z=z_1}\frac{1}{z^4+1} = -\frac{1}{4} e^{3 i\pi/4}$$
Then
$$\int_{-\infty}^\infty \frac{dx}{x^4+1} =2 \pi i (b_1+b_2) = \frac{\pi}{\sqrt{2}}$$
and finally
$$\int_0^\infty \frac{dx}{x^4+1} =  \frac{\pi}{2\sqrt{2}}$$
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