# Evaluate the integral $\int_0^\infty\dfrac{x}{\sqrt{x^5+1}}dx$

$$\int_0^\infty\dfrac{x}{\sqrt{x^5+1}}dx$$

I tried out few things, but none helped at all. Is it possible to evaluate the integral using Euler's substitutions (as demanded by my professor)? Also is there any other elementary way of solving it?

• I can't see how Euler's Substitutions can help here as the polynomial into the square root is a quintic, not a quadratic...and this integral seems to be a real nightmare, and its solution involves the Gamma function and stuff. Its primitive includes also monsters like Hypergeometric functions... Oct 30, 2019 at 18:34
• WA says it is $$\frac{2 \Gamma \left(\frac{2}{5}\right) \Gamma \left(\frac{11}{10}\right)}{\sqrt{\pi }}$$ Oct 30, 2019 at 18:34

$$B(x,y)=\int_0^{\infty}\frac{t^{x-1}}{(1+t)^{x+y}}dt$$
Where $$B(x,y)$$ is the Standard Beta function and $$\displaystyle B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ where $$\Gamma(z)$$ is the Complete Gamma function.
In your given question, what happens if you substitute $$\displaystyle x=t^{\frac{1}{5}}$$?