$\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.
 A: 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:
$$ \int_{0}^{+\infty}\frac{x^{1/3}}{(x+a)^2}\,dx = \frac{2\pi}{3\sqrt{3}\,a^{2/3}}.$$
A: Let us assume $A^3=a, B^3=b$, for simplicity. Now make a substitution $x=t^3$ which will transform the integral like this $$I=\int_{0}^\infty \frac {3t^3dt}{(t^3+A^3)(t^3+B^3)}$$ Now break this into partial fractions like this 
$$I=3\int_{0}^\infty [\frac {1}{(t^3+B^3)}-\frac {A^3}{B^3-A^3}(\frac {1}{t^3+A^3}-\frac {1}{t^3+B^3})]$$ 
Now, by using this
$$I_1=\int_{0}^\infty \frac{dt}{t^3+A^3}=\frac {2\pi}{3^{\frac{3}{2}}A^2}$$ 
(You can compute this integral very easily. Just put $x=At$ and $t=\frac {1}{p}$ and after making these two substitutions, add both integral, you will see that a quadratic is left in denominator. To remove that put $p-\frac {1}{2}=\lambda$. After making this substitution, you will easily get the value of above integral.)
Now using $I_1$, value of the $I$ is equal to 
$$I=\frac {2\pi}{3^{1/2}}[\frac{1}{B^2}-\frac {A^3}{B^3-A^3}(\frac{1}{A^2}-\frac{1}{B^2})]$$ which after simplication equal to 
$$I=\frac{2\pi}{\sqrt3 \times(B^2+A^2+AB))}$$
where $A^3=a, B^3=b$
A: You may first use $x=t^3$ and $dx=3t^2dt$, and then make the integral as the sum of two integral using the following:
$$ \frac{ 3t^3}{(t^3+a)(t^3+b)} = \frac{ \frac{-3a}{b-a}}{(t^3+a)}+ \frac{ \frac{3b}{b-a}}{(t^3+b)}$$
