<s>[INST] Evaluate $\int \frac{1}{1+x^4} dx$ 
Evaluate
$$\int_0^\infty \frac{1}{1+x^4} dx$$

It seems to me that the most natural way would be to use contour integral. The integral on the semi circle will disappear. The only thing I'm not sure about is how to find the residue. Four roots of $1+z^4 = 0$ can be found easily and they are
$$e^{\theta i} \text{ where } \theta = \frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$$
Let four solutions be $z_1,z_2,z_3,z_4$ respectively, then
$$\frac{1}{1+x^4}=\frac{1}{(z - z_1)(z-z_2)(z-z_3)(z-z_4)}$$
What would be the best way to find the residue at two points in the upper half plane, except just plug in and multiply a lot?
 [/INST] Let $x^{4}=z$
\begin{align}
\int\limits_{0}^{\infty} \frac{1}{1+x^{4}} dx 
&= \frac{1}{4} \int\limits_{0}^{\infty} \frac{z^{-3/4}}{1+z} dz \\
&= \frac{1}{4} \mathrm{B}\left(\frac{1}{4}, \frac{3}{4}\right) \\
&= \frac{\Gamma(1/4)\Gamma(3/4)}{4\Gamma(1)} \\
&= \frac{\pi}{2\sqrt{2}}
\end{align}
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