This is from Edwards Classic Treatise on the Integral Calculus, Vol 1, p167 and appears quite simple but several pages later I get stuck on partial fraction decomposition.
Show that
$$\int_{0}^{\infty}\frac{x}{1+x^5} dx = \frac{4\pi}{5\sqrt{10+2\sqrt{5}}}$$
Using the substitution $u=x^5$ gives $$\begin{align} \int_{0}^{\infty}\frac{x}{1+x^5}\mathrm{d}x &=\frac15\int_0^\infty\frac{u^{-3/5}}{1+u}\mathrm{d}u\\ &=\frac15B\left(\frac25,\frac35\right)\\ &=\frac15\cdot\frac{\Gamma\left(\frac25\right)\Gamma\left(\frac35\right)}{\Gamma\left(\frac25+\frac35\right)}\\ &=\frac{\Gamma\left(\frac25\right)\Gamma\left(1-\frac25\right)}5\\ &=\frac{\pi}{5\sin{\left(\frac{2\pi}5\right)}}\\ &=\frac{4\pi}{5\sqrt{10+2\sqrt{5}}}\\ \end{align}$$ Where I have used two representations of the Beta function, Euler's reflection formula and then the value of $\sin{(2\pi/5)}$ which is calculated here.
