Integrate $1/(x^5+1) $from $0$ to $\infty$? How can I calculate the integral $\displaystyle{\int_{0}^{\infty} \frac{1}{x^5+1} dx}$?

The change of variable $\frac{1}{1+x^{5}}=u$ gives $dx=-\frac{u^{-1-\frac{1}{5}}(1-u)^{-\frac{4}{5}}}{5}du$
Hence our integral reduces to $$\frac{1}{5}\int_{0}^{1}{u^{\frac{4}{5}-1}(1-u)^{\frac{1}{5}-1}du}=\frac{1}{5}\beta(1-\frac{1}{5},\frac{1}{5})=\frac{1}{5}\Gamma(1-\frac{1}{5})\Gamma(\frac{1}{5})=\frac{1}{5}\frac{\pi}{\sin(\frac{\pi}{5})}$$ we can actually simplify things a bit more by considering the following identity, deduced by De'Moivres formula $$\sin(5x)=16\sin^{5}(x)-20\sin^{3}(x)+5\sin(x)$$ pluggin in $x=\frac{\pi}{5}$ combined with the fact that $\sin(\frac{\pi}{5})>0$ We get $$16\sin^{4}(\frac{\pi}{5})-20\sin^{2}(\frac{\pi}{5})+5=0$$ which we can begin by solving for $\sin^{2}(\frac{\pi}{5})$ and then taking the positive root. We will end up with $\sin(\frac{\pi}{5})=\sqrt{5-\sqrt{5}}$ hence your integral will finally evaluate at $$\frac{1}{5}\frac{\pi}{\sqrt{5-\sqrt{5}}}$$
