Calculate an integral using contour integration I have to calculate the following integral using contour integration: $$\int_0^1 \frac{dx}{(x+2)\sqrt[3]{x^2(x-1)}}$$
I've tried to solve this using the residue theorem, but I don't know how to calculate the residue of the function $$f(z) = \frac{1}{(z+2)\sqrt[3]{z^2(z-1)}}$$ Then I tried to make a substitution in the real integral, so that I would get a function whose residue I know how to calculate, but I couldn't figure out what substitution would do the trick. I would really appreciate if someone could help.
 A: I will instead compute
$$ I = \int_{0}^{1} \frac{\mathrm{d}x}{(x+2)\sqrt[3]{x^2\bbox[color:red;padding:3px;border:1px dotted red;]{(1-x)}}}. $$
You will have no problem converting this to your case, depending on which branch of $\sqrt[3]{\,\cdot\,}$ is used.

1st Solution. Let $\sqrt[3]{z} = \exp(\frac{1}{3}\log z)$ be the principal complex cube root. Also, let $f(z)$ be the holomorphic function defined on $\mathbb{C} \setminus [0, 1]$ by
$$ f(z) = \frac{1}{(z+2) z \sqrt[3]{1 - z^{-1}}}. $$
Then consider the integral
$$ J = \int_{|z|=R_0} f(z) \, \mathrm{d}z, $$
where $R_0 > 2$ so that $|z| = R_0$ encloses all the singularities of $f$. Now we will compute $J$ in two ways. On one hand, by noting that $|f(z)| = \mathcal{O}(|z|^{-2})$, we get
$$ J = \lim_{R\to\infty} \int_{|z|=R} f(z) \, \mathrm{d}z = 0. $$
On the other hand, by "shrinking" the contour $|z| = R_0$ (blue circle in the figure below), we obtain a small circle around the pole $-2$ of $f$ and the dogbone contour around $[0, 1]$:

In this limit, noting that $|f(z)| = \mathcal{O}(|z|^{-2/3})$ as $z \to 0$ and $|f(z)| = \mathcal{O}(|z-1|^{-1/3})$ as $z \to 1$, we obtain
$$ J = 2\pi i \mathop{\mathrm{Res}}_{z=-2} f(z) + (e^{i\pi/3} - e^{-i\pi/3}) I. $$
In this step, we utilized the observation that, for $0 < x < 1$,
\begin{align*}
\lim_{\varepsilon \to 0^+} \sqrt[3]{1-\frac{1}{x+i\varepsilon}} &= e^{i\pi/3} \sqrt[3]{\frac{1-x}{x}}, \\
\lim_{\varepsilon \to 0^+} \sqrt[3]{1-\frac{1}{x-i\varepsilon}} &= e^{-i\pi/3} \sqrt[3]{\frac{1-x}{x}}.
\end{align*}
Finally, since $J = 0$, solving the above equality for $I$ gives
$$ I
= -\frac{2\pi i}{e^{i\pi/3} - e^{-i\pi/3}} \left( \mathop{\mathrm{Res}}_{z=-2} f(z) \right)
= \frac{\pi}{\sin(\pi/3)} \frac{1}{\sqrt[3]{12}} $$

2nd Solution. The integrand has two branch points, namely $0$ and $1$. So it would be easier if we can send one to $\infty$. This can be done, for example, by invoking the substitution
$$ t = \frac{x}{1-x}, \qquad \text{i.e.,} \qquad x = \frac{t}{1+t}. $$
Indeed, the above substitution yields
$$ I = \int_{0}^{\infty} \frac{\mathrm{d}t}{t^{2/3}(3t+2)}. $$
Now this integral can be tackled by a fairly standard manner. For example, choosing the branch cut of $\log$ as $[0, \infty)$ and using the Hankel contour (or more precisely, keyhole contour followed by limit),

we get
\begin{align*}
\left(1 - \frac{1}{e^{4\pi i/3}} \right) I
&= \int_{\text{Hankel}} \frac{1}{z^{2/3}(3z+2)} \, \mathrm{d}z \\
&= 2\pi i \left( \mathop{\mathrm{Res}}_{z=-2/3} \frac{1}{z^{2/3}(3z+2)} \right)
= \frac{2\pi i}{3 e^{2\pi i/3}(2/3)^{2/3}}.
\end{align*}
Solving this for $I$ gives the same answer.
A: I will solve the more general integral
$$I(z):=\int_0^1\frac{dx}{x^{2/3}(1-x)^{1/3}(x+z)}$$
which is found to be given by
$$I(z)=\frac{2\pi}{\sqrt{3}}z^{-2/3}(1+z)^{-1/3}$$
Note that no matter your choice of branch of the cubic root function, your integral will be proportional to $I(2)$ (specifically $-I(2)$ or $e^{i\pi/3}I(2)$ for the two most common definitions of the function as mentioned in the comments)
To prove the statement, perform the change of variables $x \to 1/x$
$$I(z)=\int_1^\infty\frac{du}{(u-1)^{1/3}(1+zu)}$$
Now it is easy to see that the integral will become elementary by setting $t=(u-1)^{2/3}$ which yields the form
$$I(z)=\frac{3}{2}\int_0^{\infty}\frac{dt}{zt^{3/2}+z+1}=\frac{3}{2z^{2/3}(1+z)^{1/3}}\int_0^{\infty}\frac{da}{a^{3/2}+1}$$
The remaining integral is standard and can be done using complex analysis, with an appropriate contour (hint below) for the result advertised in equation 2.
Hint:

 Use a pizza slice contour centered at the origin of angle $2\pi/r$. $$\int_0^\infty\frac{da}{a^r+1}=\frac{\pi}{r\sin\pi/r}, r\in \mathbb{R}$$

