$\int_0^\infty \frac{ x^{1/3}}{(x+a)(x+b)} dx $ $$\int_0^\infty \frac{ x^{1/3}}{(x+a)(x+b)} dx$$ where $a>b>0$
What shall I do?
I have diffucty when I meet multi value function.

If we apply the substitution $x=y^3$ we get:
$$ I(a,b) = 3\int_{0}^{+\infty}\frac{y^3\,dy}{(y^3+a)(y^3+b)}=\frac{3}{a-b}\int_{0}^{+\infty}\frac{a\,dy}{y^3+a}-\frac{3}{a-b}\int_{0}^{+\infty}\frac{b\,dy}{y^3+b}$$
but if we set, for any $c>0$,
$$ J(c) = \int_{0}^{+\infty}\frac{c\,dy}{y^3+c},$$
we simply have $J(c)= c^{1/3}\,J(1)$ through the substitution $y=c^{1/3} z$. At last,
$$ J(1) = \int_{0}^{+\infty}\frac{dy}{y^3+1}=\int_{0}^{1}\frac{dy}{y^3+1}+\int_{0}^{1}\frac{y\,dy}{y^3+1}=\int_{0}^{1}\frac{dy}{1-y+y^2}=\frac{2\pi}{3\sqrt{3}},$$
hence:

$$ \int_{0}^{+\infty}\frac{x^{1/3}\,dx}{(x+a)(x+b)}=\color{red}{\frac{2\pi}{\sqrt{3}}\cdot\frac{a^{1/3}-b^{1/3}}{a-b}}.$$

By considering the limit of the RHS as $b\to a$, we also have:
$$ \int_{0}^{+\infty}\frac{x^{1/3}}{(x+a)^2}\,dx = \frac{2\pi}{3\sqrt{3}\,a^{2/3}}.$$
