Evaluate this integral $$\int_0^1(x^2-x^3)^{-\frac 1 3}dx$$ by using a barbell-shaped contour with shrinking end surrounding 0 and 1, together with a large circle containing the barbell.
The branch cut connects 0 and 1. Let the value of $f(z)=z^{-\frac 2 3}(1-z)^{-\frac1 3}$ over the branch cut be positive. When a point over the branch cut moves along the red curve to the corresponding point under the branch cut, $\Delta \arg(z^{-\frac 2 3})=e^{-i{\frac {3\pi}{4}}}$ and $\Delta \arg\big((1-z)^{-\frac1 3}\big)=0$. So $e^{-i\frac {4\pi} 3}$ should be multiplied under the branch cut.
The integral over the outer circle equals:$$-2\pi iRes(f(z),\infty)=2\pi iRes(f(\frac 1 t) \cdot \frac 1 {t^2},0)=2\pi iRes(\frac 1 t (t-1)^{-\frac 1 3})=2\pi i(-1)^{-\frac 1 3}.$$ Because of the branch we chose, $(-1)^{-\frac 1 3}=e^{-i{\frac \pi 3}}.$ We get $$(1-e^{-i{\frac {4\pi} 3}})\int_0^1(x^2-x^3)^{-\frac 1 3}dx + 2\pi ie^{-i{\frac \pi 3}}=0.$$ So,$$\int_0^1(x^2-x^3)^{-\frac 1 3}dx=\frac {2\pi} {\sqrt 3}.$$