Evaluate $\int_0^1 \frac{dx}{(x^2-x^3)^\frac13}$ using contour integration
I try to proceed by first calculating residues of $$f(x) = \frac{1}{(x^2-x^3)^\frac13} = \frac{1}{x\left(\frac1x-1\right)^\frac13}$$
Then answer to integral can be provided by Cachy's Residue theorem = $2i\pi(\text{Residue at } x=0 +\text{ Residue at } x=1)$
Above function has simple pole at $x=0$ but I am unable to find out what kind of singularity $f$ has at $x= 1$.
Is this the right approach? Or is there any way to expand given function(Laurent's series) to find out all residues at once?