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
 A: Using the very good method suggested and recommended by   Mhenni Benghorbal, the problem becomes effectively quite simple. Using $$1+x^4=\frac{1}{t}$$ we then have $$x=\frac{\sqrt[4]{1-t}}{\sqrt[4]{t}}$$ $$dx=-\frac{1}{4 (1-t)^{3/4} t^{5/4}}$$ $$\int_0^\infty \frac{dx}{(x^4+1)^5}=\frac{1}{4} \int_0^1 t^{15/4} (1-t)^{-3/4} ~dt =\frac{1}{4} B\left(\frac{19}{4},\frac{1}{4}\right) $$ The final result can be written in many different forms and in particular in terms of $\Gamma$ as you posted; in fcat, the result simplifies to $\frac{1155 \pi }{4096 \sqrt{2}}$.
More generally, still using the same technique, you could easily establish $$\int_0^\infty \frac{dx}{(x^m+1)^n}=\frac{\Gamma \left(1+\frac{1}{m}\right) \Gamma \left(n-\frac{1}{m}\right)}{\Gamma
   (n)}$$ provided $\Re(m n)>1\land \Re(m)>0$.
Going even further,$$\int_0^\infty \frac{x^p~~dx}{(x^m+1)^n}=\frac{\Gamma \left(\frac{p+1}{m}\right) \Gamma \left(\frac{m n-p-1}{m}\right)}{m
   \Gamma (n)}$$ provided $\Re(m n-p)>1\land \Re(p)>-1\land \Re(m)>0$.
I should underline that, for your specific problem, trying to compute first the antiderivative would have been a small nightmare (a CAS allowed to generate a quite long expression).
A: 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 $I = \dfrac{1}{2}\cdot \dfrac{1}{2}B(1/4,19/4) = \dfrac{1}{4}\dfrac{\Gamma(1/4)\Gamma(19/4)}{\Gamma(5)}$
A: The result is indeed true, and the method used to derive it has already been described. Now all that's left is to simplify it by using Euler's reflection formula for the $\Gamma$ function, in order to finally arrive 
at the desired result, $~I=\dfrac{1155}{2^{12}}\cdot\dfrac\pi{\sqrt2}$
A: \begin{align}
x^4+1 & = (x^4 +2x^2 + 1)-2x^2 \\[8pt]
& = (x^2+1)^2 - (\sqrt{2}\, x)^2 \\[8pt]
& = (x^2 - \sqrt{2}\, x +1)(x^2 + \sqrt{2}\, x + 1)
\end{align}
Then go to partial fractions.
