Express $\int_0^1\frac{dt}{t^{1/3}(1-t)^{2/3}(1+t)}$ as a closed path integral enclosing the interval $(0,1)$ From an old complex analysis qualifier:

Define $$I=\int_0^1\frac{dt}{t^{1/3}(1-t)^{2/3}(1+t)}.$$ 
  
  
*
  
*Express $I$ as a closed path integral enclosing the interval
  $(0,1)$.
  
*Evaluate $I$.

Ideas: At first I thought this was an exercise in the Schwarz-Christoffel integral, but now I'm not so sure. I'm thinking I'm just not aware of the method they're suggesting, and I was hoping someone could point me to it.
Thanks

Consider $$f(z) = e^{-1/3\mathrm{LogA} z} e^{-2/3\mathrm{LogB} (1-z)} \frac{1}{1+z}$$
where $\mathrm{LogA}$ is the logarithm with branch cut on the negative real axis and argument from $-\pi$ to $\pi$ and $\mathrm{LogB}$ has the branch cut on the positive real axis and argument from $0$ to $2\pi.$
Then we have continuity of $f(z)$ across the negative real axis because from above we have
$$e^{-1/3\mathrm{LogA} z} e^{-2/3\mathrm{LogB} (1-z)} 
= e^{-1/3\log x - 1/3\times\pi i} e^{-2/3\log(1-x)-2/3\times 2\pi i} $$
and from below
$$e^{-1/3\mathrm{LogA} z} e^{-2/3\mathrm{LogB} (1-z)} 
= e^{-1/3\log x + 1/3\times\pi i} e^{-2/3\log(1-x)}$$
because $$e^{- (1/3+4/3)\times\pi i} = e^{ 1/3\times\pi i}.$$
In this continuity argument we have used $x$ to denote the absolute value of $z$ on the negative real axis.
It follows by Morera's Theorem that $f(z)$ is actually analytic across the part of the cut on the negative real axis and $f(z)$ only has a cut on $[0,1].$
Now considering the counterclockwise handle-shaped contour enclosing $(0,1)$ we get that the integral above the cut is
$$- e^{-2/3\times 2\pi i} I$$
where $I$ is the desired value and
below the cut we get the value $I.$
Now the residue of $f(z)$ at $z=-1$ is
$$e^{-1/3\times \pi i} e^{-2/3 \log 2 - 2/3\times 2\pi i}
= 2^{-2/3} e^{-5/3\pi i}.$$
We thus obtain
$$I = - 2\pi i\frac{2^{-2/3} e^{-5/3\pi i}}{1-e^{-2/3\times 2\pi i}}
= -2\pi i 2^{-2/3} \frac{e^{-\pi i}}{e^{2/3\times \pi i}-e^{-2/3\times\pi i}}\\
= \pi  2^{-2/3} \frac{2i}{e^{2/3\times \pi i}-e^{-2/3\times\pi i}}
= \pi  \frac{2^{-2/3}}{\sin(2/3\pi)}
= \pi\frac{\sqrt[3]{2}}{\sqrt{3}}.$$
There is no residue at infinity to consider because $f(z)$ is $O(1/R^2)$ there on a ray to infinity at distance $R$ from the origin.
This technique is documented at Wikipedia. In the present case we get that
$$-\lim_{|z|\to\infty} z f(z) \sim -\lim_{|z|\to\infty} \frac{z}{|z|^2} = 0.$$
This concludes the argument.
