I have the following integral to solve:
$$\int_0^\infty \frac{\sqrt[3]{x+1} - \sqrt[3]{x}}{\sqrt{x}} \, \mathrm dx$$
I tried substitution $u= x+1$ and $u=\sqrt[3]{x+1}$ then partial integration but that didn't really help because it didn't become any simpler.
Thank you.
Hint. If you know the Euler integral (beta function) $$ \int_{0}^{1}u^{a−1}(1−u)^{b−1}dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} \tag1 $$ then you may perform a change of variable $\displaystyle u=\frac{1}{1+x}\,$, $\displaystyle x=\frac{1}{u}-1$, in your initial integral to obtain $$ \begin{align} \int_0^\infty \frac{\sqrt[3]{x+1} - \sqrt[3]{x}}{\sqrt{x}} \, \mathrm dx &=\int_0^1 \frac{(1/u)^{\frac13}-(1/u-1)^{\frac13}}{(1/u-1)^{\frac12}} \frac{1}{u^2} \mathrm du\\\\ &=\int_0^1 \frac{1-(1-u)^{\frac13}}{(1-u)^{\frac12}} u^{-11/6} \mathrm du. \end{align} $$ Now, observe that using $(1)$ we get, for $s>0$, $$ \begin{align} \int_0^1 \frac{1-(1-u)^{\frac13}}{(1-u)^{\frac12}} u^{s} \mathrm du&=\frac{\Gamma(1/2)\:\Gamma(1+s)}{\Gamma(3/2+s)}-\frac{\Gamma(5/6)\:\Gamma(1+s)}{\Gamma(11/6+s)} \end{align} \tag2 $$ and since both sides of $(2)$ are analytic for $\Re s>-2$ (the singulariy of the integrand at $u=0^+$ is integrable as soon as $\Re s>-2$) then the identiy can be extended to this case allowing one to write $$ \begin{align} \int_0^1 \frac{1-(1-u)^{\frac13}}{(1-u)^{\frac12}} u^{-11/6} \mathrm du &=\lim_{s \to -11/6}\left(\frac{\Gamma(1/2)\:\Gamma(1+s)}{\Gamma(3/2+s)}-\frac{\Gamma(5/6)\:\Gamma(1+s)}{\Gamma(11/6+s)}\right)\\\\ &=\lim_{s \to -11/6}\Gamma(1+s)\left(\frac{\sqrt{\pi}\:}{\Gamma(3/2+s)}-\frac{\Gamma(5/6)\:(11/6+s)}{\Gamma(17/6+s)}\right)\\\\ &=\frac{2\sqrt{\pi}\:\Gamma(1/6)}{5\Gamma(2/3)}, \end{align} $$ with some algebra, consequently
$$ \int_0^\infty \frac{\sqrt[3]{x+1} - \sqrt[3]{x}}{\sqrt{x}} \, \mathrm dx =\frac{2\sqrt{\pi}\:\Gamma(1/6)}{5\Gamma(2/3)}. $$
