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

My approach is similar to Marko Riedel's approach.
Consider $$f(z) = \frac{1}{z^{1/3} (z-1)^{2/3} (1+z)} = \frac{1}{|z|^{1/3} e^{i/3 \arg z}|z-1|^{2/3} e^{2/3i \arg(z-1)} (1+z)}$$
where $0 \le \arg(z), \arg(z-1) < 2 \pi$.
By omitting the line segment $[0,1]$, $f(z)$ is single-valued on the complex plane.
Then integrating around a dumbbell/dog-bone contour,
$$ \int_{0}^{1}\frac{dx}{x^{1/3}(1-x)^{2/3}e^{2 \pi i /3}(1+x)} + \int_{1}^{0} \frac{dx}{x^{1/3} e^{2 \pi i/3} (1-x)^{2/3}e^{2 \pi i/3}(1+x)}$$
$$ = \  2 \pi i \Big( \text{Res}[f(z),-1] + \text{Res}[f(z),\infty]\Big)$$
where
$$\begin{align} \text{Res}[f(z),-1] &= \lim_{z \to -1} \frac{1}{|z|^{1/3} e^{i/3 \arg z}|z-1|^{2/3} e^{2/3i \arg(z-1)}} \\ &= \frac{1}{e^{i \pi/3}2^{2/3}e^{2 \pi i/3}}\\ &= -2^{-2/3} \end{align}$$
and 
$$ \text{Res}[f(z),\infty] = 0$$
since the Laurent expansion of $f(z)$ at $\infty$ has no $ \displaystyle \frac{1}{z}$ term.
Therefore,
$$ - i \sqrt{3} \int_{0}^{1} \frac{dx}{x^{1/3} (1-x)^{2/3}(1+x)} = 2 \pi i \left(-2^{-2/3} \right) $$ 
which implies
$$ \int_{0}^{1} \frac{dx}{x^{1/3} (1-x)^{2/3}(1+x)} = \frac{\pi \ 2^{1/3}}{ \sqrt{3}}$$
