<s>[INST] What's the integral of $\int_0^\infty \frac{dx}{(x^4+1)^5}$ $$\int_0^\infty \frac{dx}{(x^4+1)^5}$$
My answer would be : $\dfrac{\Gamma(\tfrac{1}{4})\Gamma(\tfrac{19}{4})}{4\Gamma(5)}$
Solution: You can use this technique. –  Mhenni Benghorbal
 [/INST] Let $x = \sqrt{\tan \theta}$. Then, $dx = \dfrac{1}{2\sqrt{\tan \theta}}\cdot \sec^2 \theta d\theta =\dfrac{d\theta}{2\sin^{1/2}\theta\cos^{3/2} \theta}$. If $x = 0$, $\theta = 0$ and if $x \to \infty \ \Rightarrow \ \theta = \pi/2$. 
$I =\int_{0}^{\infty}\dfrac{dx}{(x^4 + 1)^5} = \dfrac{1}{2}\int_{0}^{\pi/2}\dfrac{\cos^{10}\theta d\theta}{\sin^{1/2}\theta\cos^{3/2}\theta} = \dfrac{1}{2}\int_{0}^{\pi/2}\sin^{-1/2}\theta\cos^{17/2}\theta d\theta$
Beta function: $B(x,y) = 2\int_{0}^{\pi/2}\sin^{2x - 1}\theta \cos^{2y - 1}\theta d\theta $
Thus, $x = 1/4$ and $y = 19/4$ and $I = \dfrac{1}{2}\cdot \dfrac{1}{2}B(1/4,19/4) = \dfrac{1}{4}\dfrac{\Gamma(1/4)\Gamma(19/4)}{\Gamma(5)}$
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