Complex integration without beta function Recently I asked you 2 integration problems to light me the way to solve other ones. That worked but in some of the problems I struggle. For example,
$$\int_0^1{dx\over (x^2+1)(x^2(1-x))^{1/3}}$$
in this question, I couldn't apply beta function, in fact I want to solve it witohut using it. How can I do it?
Thanks in advance!
 A: $\newcommand{\Res}{\operatorname{Res}}\newcommand{\Arg}{\operatorname{Arg}}$I understand that in Complex Analysis class, if the problem is  to find a definite integral, then very likely it is expected from you to use the Residue Theorem on that specific integral. 
I provided you a link in the comments, but I will still guide you to the right place. I use an approach similar to Mark Viola's approach.
Let's start by defining $$f(z):=\frac{1}{(1+z^2)(-z)^{2/3}(1-z)^{1/3}}$$ As usual I use the prinicipal value logarithm to define the powers. That is also the main reason I put $(-z)$ in the denominator which seems strange in the first place.
Okay now you can verify that $f(z)$ is meromorphic on $\mathbb{C}\setminus [0,1]$.  We use the  dog bone contour $C$ where the dog bone is on $[0,1]$. The large circular part of the contour will be called $C_R$ and the radius is then $R\gg 1$.  It is easy to see using the ML-lemma/estimation lemma that the contribution of it is zero when $R\to\infty$. We also have the contours on the dog bone, some have circular form with radius $\epsilon>0$, but even those will have no contribution when $\epsilon\to 0$. These things are all left for you to verify! We are only left with the straight lines that go from $0$ to $1$ call it $K^+$ and the other one going from $1$ to $0$ from below call it $K^-$. So:
\begin{align}
\oint_C f(z)\,dz = \int_{K^+}f(z)\,dz+\int_{K^-}f(z)\,dz
\end{align}
You can verify with the choice for the principal log that we have:
\begin{align}\tag{1}
\oint_C f(z)\,dz = (e^{i2\pi/3}-e^{-i2\pi/3})\int^1_0 \frac{1}{(1+x^2)x^{2/3}(1-x)^{1/3}}\,dx
\end{align}
In $(1)$ you have to do some work. We can simplify more to get:
\begin{align}
\oint_C f(z)\,dz = 2i\sin(2\pi/3)\int^1_0 \frac{1}{(1+x^2)x^{2/3}(1-x)^{1/3}}\,dx
\end{align}
We did not apply the Residue Theorem yet. How many residues do we have inclosed by our contour? Two, namely $z=i$ and $z=-i$. So we have:
\begin{align}
\oint_C f(z)\,dz  =2\pi i\left(\Res_{z=i} f(z)+\Res_{z=-i}f(z)\right)
\end{align}
I do one residue; the other one is for you.
\begin{align}
\Res_{z=i} f(z) &= \Res_{z=i}\frac{1}{(1+z^2)(-z)^{2/3}(1-z)^{1/3}}\\
&=\frac{1}{2i(-i)^{2/3}(1-i)^{1/3}}
\end{align}
We have $(-i)^{2/3}=\exp\left( i\frac{2}{3}\Arg(-i)\right)=\exp(-i\pi/3)$ and $(1-i)^{1/3}=\exp\left(\frac{1}{3}\ln(\sqrt[]{2})+ i\frac{1}{3}\Arg(1-i)\right)=\sqrt[6]{2}\exp(-i\pi/12)$ so:
\begin{align}
\Res_{z=i} f(z) = \frac{e^{i5\pi/12}}{2i\sqrt[6]{2}}
\end{align}
The other residue is:
\begin{align}
\Res_{z=-i} f(z) =- \frac{e^{-i5\pi/12}}{2i\sqrt[6]{2}}
\end{align}
So we get:
\begin{align}
\oint_C f(z)\,dz=\frac{2\pi i}{\sqrt[6]{2}}\sin(5\pi/12)
\end{align}
Finally:
\begin{align}
\int^1_0 \frac{1}{(1+x^2)x^{2/3}(1-x)^{1/3}}\,dx=\pi\frac{\sin(5\pi/12)}{\sqrt[6]{2}\sin(2\pi/3)}
\end{align}
That can be simplified further to get:

\begin{align}
\int^1_0 \frac{1}{(1+x^2)x^{2/3}(1-x)^{1/3}}\,dx=\frac{\pi}{\sqrt[3]{4}}\left(1+\frac{1}{\sqrt[]{3}}\right)
\end{align}

A: I am not sure there is a simple way without using Euler's Beta function... Anyway, once
$$ \int_{0}^{1}\frac{x^{2k}}{x^{2/3}(1-x)^{1/3}}\,dx = \frac{\Gamma\left(\frac{2}{3}\right)\Gamma\left(2k+\tfrac{1}{3}\right)}{\Gamma(2k+1)} \tag{A}$$
is proved (for instance by integration by parts and induction on $k$), the original integral is converted into the hypergeometric function
$$ \sum_{k\geq 0}\frac{(-1)^k\, \Gamma\left(\frac{2}{3}\right)\Gamma\left(2k+\tfrac{1}{3}\right)}{\Gamma(2k+2)}=\underbrace{\Gamma\left(\tfrac{2}{3}\right)\Gamma\left(\tfrac{1}{3}\right)}_{\frac{2\pi}{\sqrt{3}}}\;\phantom{}_2 F_1\left(\tfrac{1}{6},\tfrac{2}{3};\tfrac{3}{2};-1\right) \tag{B}$$ 
which can be obtained from the Taylor series of $(1-x)^{2/3}$ by a step of the discrete Fourier transform. Indeed:
$$\sum_{k\geq 0}\frac{x^k\, \Gamma\left(\frac{2}{3}\right)\Gamma\left(2k+\tfrac{1}{3}\right)}{\Gamma(2k+2)}=\frac{\pi\sqrt{3}}{2\sqrt{x}}\left[\left(1+\sqrt{x}\right)^{2/3}-\left(1-\sqrt{x}\right)^{2/3}\right] \tag{C}$$
and with a bit of care in managing the determinations of the square/cube roots this leads to:
$$ \int_{0}^{1}\frac{dx}{(1+x^2)\,x^{2/3}(1-x)^{1/3}}=\color{red}{\frac{\pi}{\sqrt[3]{4}}\left(1+\frac{1}{\sqrt{3}}\right)}.\tag{D}$$
Equivalently, one may invoke the Lagrange inversion theorem.
