Show that $\int_{0}^{1}\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)} dx = 2\pi\frac{\sqrt{3}}{27}$ The given question is from Complex Variables written by Levinson and Redheffer, Problem 7 in Chapter 4.  We want to show that $\int_{0}^{1}\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)}\,dx = 2\pi\frac{\sqrt{3}}{27}$. Below is my first attempt. 
Let $C$ be the a contour that consists of:
a. A counter-clockwise oriented closed circle $C_1$ centered at the point $z = 1$ with radius $\epsilon$
b. A counter-clockwise oriented closed circle $C_0$ centered at the point $z = 0$ with radius $\epsilon$
c. A directed path $P_1$that starts from point $z = \epsilon$, ends at point $z =1 - \epsilon$ and joins $C_0$ and $C_1$ and lies in the upper half of $x$ axis.
d. A directed path $P_2$ that starts from point $z = 1 - \epsilon$, ends at point $z =\epsilon$ and joins $C_0$ and $C_1$ and lies in the lower half of $x$ axis.  
Namely $C = C_1 \rightarrow P_2 \rightarrow C_0 \rightarrow P_1$. Also we know that the angle of $\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)}$ will change by $\frac{-2\,\pi\,i}{3}$ and hence on $P_2$ we have $\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)}\,e^{\large\frac{-2\,\pi\,i}{3}}$ while on $P_1$ we have just$\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)}$. I do not know if we need to find the residue at infinity but for the given function it does not exist. So I try $x^2 = (\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)} + x)\,[\sqrt[\leftroot{-3}\uproot{3}3]{x^4(1 - x)^2} - x\,\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)} + x^2]$ and try to find the contour integral of $\large\frac{x}{\sqrt[\leftroot{-3}\uproot{3}3]{x^4(1 - x)^2} + x\,\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)}] + x^2}$ around $C$. However, then I do not know how to work out the result when the denominator is $x^2$. 

Let $f(z)$ be the branch of $\sqrt[\leftroot{-3}\uproot{3}3]{z^2(1 - z)}$ that is real positive at $z$ on $P_1$. Then the value of $f(x)$ for $x>1$ is $\sqrt[\leftroot{-3}\uproot{3}3]{x^2(x-1)}\cdot e^{-\frac{\pi i}{3}}$ and  $f(x)=\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1-x)}\cdot e^{-\frac{2\pi i}{3}}$ on $P_2$. Therefore, letting $ \epsilon\to 0$, we have$$
\int_0^1 \sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)} dx+e^{-\frac{2\pi i}{3}}\int_1^0
\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1-x)}dx=2\pi i\operatorname{Res}(\infty).\tag{1}
$$
Evaluation of $\operatorname{Res}(\infty)$:
Since $\sqrt[\leftroot{-3}\uproot{3}3]{1-t}$ has its expansion 
$$
\sqrt[\leftroot{-3}\uproot{3}3]{1-t}=1-\frac{t}{3}-\frac{t^2}{9}-\frac{5t^3}{81}-\cdots \quad(|t|<1), 
$$
the expansion of $f(x)$ for $1<x<\infty$
is
\begin{align*}
f(x)&=\sqrt[\leftroot{-3}\uproot{3}3]{x^2(x-1)}\cdot e^{-\frac{\pi i}{3}}\\
&=e^{-\frac{\pi i}{3}}\cdot x\left(1-\frac{1}{3x}-\frac{1}{9x^2}-\frac{5}{81x^3}-\cdots \right).
\end{align*} 
Thus we have $$\operatorname{Res}(\infty)=\frac{1}{9}e^{-\frac{\pi i}{3}}.$$
Recall that if $f(z)=\sum_{n=-\infty}^\infty \frac{c_n}{z^n} \,(r<|z|<\infty)$, then $\operatorname{Res}(\infty)=-c_1.$
From $(1)$ we have \begin{align*}
&\left(1-e^{-\frac{2\pi i}{3}}\right)\int_0^1 \sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)} dx
=\frac{2\pi i}{9}e^{-\frac{\pi i}{3}},\\
&\int_0^1 \sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)} dx=2\pi\frac{\sqrt{3}}{27}.
\end{align*}
