Using the residue calculus, evaluate the integral:
$\int_{0}^{\infty} \frac{ln(x)}{x^{1/3}(x^2-1)} dx$
I know, that the function has singularities at +1, -1, 0. I find the rule which says that:
$\int_{0}^{\infty} f(x) ln(x) dx = -\frac{1}{2}Re \sum\limits_{C-[0;{\infty})} res_z(f(z)log_{2\pi}^{2}z)$
I need help at the moment because I don't get the right result and I don't know where I could have made a mistake and if this rule can be used.