<s>[INST] Complex integration $\int_{0}^{1}\frac{\sqrt[3]{4x^{2}\left(1-x\right)}}{\left(1+x\right)^{3}}dx$ I have integral
$$I_1=\int_{0}^{1}\frac{\sqrt[3]{4x^{2}\left(1-x\right)}}{\left(1+x\right)^{3}}dx.$$
I tried something like this:
$$f\left(z\right)=\frac{\sqrt[3]{4z^{2}\left(1-z\right)}}{\left(1+z\right)^{3}}=\frac{\left(1-z\right)}{\left(1+z\right)^{3}}\left(\frac{2z}{1-z}\right)^{\frac{2}{3}}$$

Using the figure I have
$$I=\int_{C_{R}}f\left(z\right)dz+\int_{C_{r1}}f\left(z\right)dz+\int_{z_{1}}f\left(z\right)dz+\int_{C_{r2}}f\left(z\right)dz+\int_{z_{2}}f\left(z\right)dz.\tag1$$
Since I have only one singularity in my contour:
$$\operatorname {Res}\left(f,z=-1\right) =\frac{1}{\left(3-1\right)!}\cdot\lim_{z\rightarrow\left(-1\right)}\left[\left(z+1\right)^{3}\cdot f\left(z\right)\right]^{''}\\
 =\frac{1}{2}\cdot\lim_{z\rightarrow\left(-1\right)}\frac{-2\cdot4^{\frac{1}{3}}z^{2}}{9\left(-z^{2}\left(-1+z\right)\right)^{\frac{5}{3}}}\\
 =\frac{1}{2}\cdot\frac{-2\cdot4^{\frac{1}{3}}\left(-1\right)^{2}}{9\left(-\left(-1\right)^{2}\cdot\left(-1-1\right)\right)^{\frac{5}{3}}}\\
 =-\frac{2^{\frac{2}{3}}\cdot1}{9\left(2\right)^{\frac{5}{3}}}=-\frac{1}{9}2^{\frac{2}{3}-\frac{5}{3}}=-\frac{1}{9}2^{-1}=-\frac{1}{18}.$$
Thus I got
$$I=2\pi i\cdot\left(-\frac{1}{18}\right)=-\frac{\pi i}{9}.$$
Now, I use Jordan's lemma and get:
$$\lim_{z\rightarrow0}z\cdot f\left(z\right)=0$$
$$\lim_{z\rightarrow1}\left(z-1\right)f\left(z\right)=0$$
$$\lim_{z\rightarrow\infty}z\cdot f\left(z\right)=0$$
So I conclude that
$$\lim_{r\rightarrow0}\int_{C_{r1}}f\left(z\right)dz=0$$
$$\lim_{r\rightarrow0}\int_{C_{r2}}f\left(z\right)dz=0.$$
$$\lim_{R\rightarrow\infty}\int_{C_{R}}f\left(z\right)dz=0.$$
When I put $r\rightarrow0$ and $R\rightarrow\infty$ into (1) and put that on $z_1$ the argument of the function  is 0, and on $z_2$ the argument is $2\pi \cdot \frac{2}{3}$ then I have:
$$I_{1}\left(1-e^{i\frac{4\pi}{3}}\right)=-\frac{\pi}{9}i.$$
Now I have the problem because on the left side I have real and imaginary part and on the right hand side I only have imaginary part and I don't know what to do.
 [/INST] You already showed that the integral over the three circles vanishes in the limit, and we are left with (since the argument of $1 - z$ increases by $-2\pi$ when moving from the upper to the lower edge)
$$I_{1}\left(1 - e^{-2\pi i / 3}\right) = 2\pi i\operatorname{Res}(f,-1).$$
Since we are dealing with the third root, the difficult part will be to get the argument right.
Write
$$g(z) = 4^{1/3}z^{2/3}(1 - z)^{1/3}$$
so
$$f(z) = \frac{g(z)}{(1 + z)^3}$$
and
$$\operatorname{Res}(f,-1) = \frac12 g''(-1).$$
Now our definition of $f$ determines how $g$ is defined on this branch, namely in such a way that for $x \in (0,1)$ we have that $g(x)$ is positive and real. In order to be able to determine the branch of its derivatives, we have to express those in $g$ itself and we see:
$$g'(z) = \left(\frac2{3z} - \frac1{3(1 - z)}\right)g(z) = q(z)g(z)$$
if we set
$$q(z) = \frac2{3z} - \frac1{3(1 - z)},$$
hence also
$$g''(z) = q'(z)g(z) + q(z)g'(z) = \left(q'(z) + q(z)^2\right)g(z).$$
Let's write out $q'(z)$:
$$q'(z) = -\frac2{3z^2} - \frac1{3(1 - z)^2}.$$
Then
$$g''(-1) = \left(q'(-1) - q(-1)^2\right)g(-1) = \left(-\frac56 - \left(-\frac34\right)^2\right)g(-1) = -\frac{g(-1)}{18}.$$
We clearly have that $|g(-1)| = 2$, so we only need its argument. The argument of $g(x)$ for $x \in (0,1)$ is 0, and when going from $x$ to $-1$ the argument of $1 - z$ doesn't change, while that of $z$ goes from 0 to $\pi$. It follows that the argument of $g(z)$ goes up by $\frac23\pi$, so
$$g''(-1) = -\frac{e^{2\pi i/3}}9$$
and
$$\operatorname{Res}(f,-1) = -\frac{e^{2\pi i/3}}{18}.$$
Finally
$$I_{1} = 2\pi i\frac{\operatorname{Res}(f,-1)}{1 - e^{-2\pi i / 3}} = -\frac{2\pi i}{18}\frac{e^{2\pi i/3}}{1 - e^{-2\pi i / 3}} = \frac{\pi\sqrt3}{27}.$$
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