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

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$\ds{I \equiv \int_{0}^{1}{\dd t \over t^{1/3}\pars{1 - t}^{2/3}\pars{1 + t}}:\ {\large ?}}$

\begin{align}
I&=\int_{\infty}^{1}{-\,\dd t/t^{2} \over t^{-1/3}\pars{1 - 1/t}^{2/3}\pars{1 + 1/t}}
=\int^{\infty}_{1}{\dd t \over \pars{t - 1}^{2/3}\pars{t + 1}}
=\int^{\infty}_{0}{t^{-2/3} \over t + 2}\,\dd t
\\[3mm]&=2^{-2/3}\int^{\infty}_{0}{t^{-2/3} \over t + 1}\,\dd t\tag{1}
\end{align}

Integrate
$\ds{\int_{C}{z^{-2/3} \over z + 1}\,\dd z}$ where $C$ is a 'key-hole contour' and $z^{-2/3} = \verts{z}^{-2/3}\expo{-2\phi\pars{z}\ic/3}$ with $z \not= 0$ and $0 < \phi\pars{z} < 2\pi$:
\begin{align}
2\pi\ic\expo{-2\pi\ic/3}
&=
\int^{\infty}_{0}{t^{-2/3} \over t + 1}\,\dd t
+
\int_{\infty}^{0}{t^{-2/3}\expo{-2\pars{2\pi}\ic/3} \over t + 1}\,\dd t
=
\pars{1 - \expo{-4\pi\ic/3}}\int^{\infty}_{0}{t^{-2/3} \over t + 1}\,\dd t
\\[3mm]\mbox{Then,}&
\int^{\infty}_{0}{t^{-2/3} \over t + 1}\,\dd t
=
2\pi\ic\,{\expo{-2\pi\ic/3} \over 1 - \expo{-4\pi\ic/3}}
=
{2\pi\ic \over \expo{2\pi\ic/3} - \expo{-2\pi\ic/3}}
=
{\pi \over \sin\pars{2\pi/3}}
=
{\pi \over \root{3}/2}
\\[3mm]&={2\root{3} \over 3}\,\pi
\end{align}
By replacing this result in $\pars{1}$, we find:
$$\color{#0000ff}{\large%
I=\int_{0}^{1}{\dd t \over t^{1/3}\pars{1 - t}^{2/3}\pars{1 + t}}
=
{2^{1/3}\root{3} \over 3}\,\pi}
$$
