$\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.

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 $A^3=a, B^3=b$
