Evaluate $\int_{0}^{1}\frac{\sqrt[4]{x (1-x)^{3}}}{(1+x)^{3}}\mathrm{d}x$ How to evaluate the two integrals below:
$$\int_{0}^{1}\frac{\sqrt[4]{x\left ( 1-x \right )^{3}}}{\left ( 1+x \right )^{3}}\mathrm{d}x$$
and
$$\int_{0}^{1}\frac{\sqrt[3]{x\left ( 1-x \right )^{2}}}{\left ( 1+x \right )^{3}}\mathrm{d}x$$
After a long time thinking,I still in trouble.
 A: Using the definition of Beta function
$$B\left ( p,q \right )=\frac{\Gamma \left ( p \right )\Gamma \left ( q \right )}{\Gamma \left ( p+q \right )}=\int_{0}^{1}x^{p-1}\left ( 1-x \right )^{q-1}\mathrm{d}x~ ~ ~ ~ \left ( \Re q>0,\Re p>0 \right )$$
In general,
$$I\left ( m,n \right )=\int_{0}^{1}\frac{\sqrt[n]{x^{m}\left ( 1-x \right )^{n-m}}}{\left ( 1+x \right )^{3}}\mathrm{d}x$$
So in order to translate your integral into Beta function,you may perform a change of variable like this:
$$t=\frac{2x}{1+x},~ ~ ~ 1-t=\frac{1-x}{1+x},~ ~ ~ \mathrm{d}t=\frac{2\mathrm{d}x}{\left ( 1+x \right )^{2}}$$
Hence we have,
\begin{align*}
\int_{0}^{1}\frac{\sqrt[n]{x^{m}\left ( 1-x \right )^{n-m}}}{\left ( 1+x \right )^{3}}\mathrm{d}x &=\int_{0}^{1}\left ( \frac{x}{1+x} \right )^{\frac{m}{n}}\left ( \frac{1-x}{1+x} \right )^{\frac{n-m}{n}}\frac{\mathrm{d}x}{\left ( 1+x \right )^{2}} \\
 &=2^{-\frac{n+m}{n}}\int_{0}^{1}t^{\frac{m}{n}}\left ( 1-t \right )^{\frac{n-m}{n}}\mathrm{d}t \\
 &=\frac{2^{-\frac{n+m}{n}}}{\Gamma \left ( 3 \right )}\Gamma \left ( \frac{m+n}{n} \right )\Gamma \left ( \frac{2n-m}{n} \right ) \\
 &=2^{-\frac{2n+m}{n}}\cdot \frac{m}{n}\cdot \frac{n-m}{n}\cdot \Gamma \left ( \frac{m}{n} \right )\cdot \Gamma \left ( 1-\frac{m}{n}\right )\\
 &=2^{-\frac{2n+m}{n}}\cdot \frac{m\left ( n-m \right )}{n^{2}}\cdot \frac{\pi }{\sin\left ( \frac{m\pi }{n} \right )}
\end{align*}
In the particular case of $m=1,n=4~and~3$,we can easily get
$$\Large\color{blue}{I\left ( 1,4 \right )=\frac{3\sqrt[4]{2}}{64}\pi }$$
$$\Large\color{blue}{I\left ( 1,3\right )=\frac{\pi }{18}\frac{\sqrt[3]{4}}{\sqrt{3}}}$$
A: As @tired mentioned in the comments, these integrals may be attacked via complex analysis.  I will illustrate the first one - the second one is quite similar in execution.  We use a contour that avoids the branch cuts along $(-\infty,0]$ and $(-\infty,1]$.
We consider the contour integral
$$\oint_C dz \frac{z^{1/4} (z-1)^{3/4}}{(z+1)^3} $$
where $C$ is the following contour: (almost...imagine the middle hump at the origin)

Note that the outer arc has radius $R$ and the small arcs have radius $\epsilon$.  In the limit as $\epsilon \to 0$, the contour integral is equal to
$$e^{i \pi}PV \int_{R}^0 dx \frac{e^{i \pi/4} x^{1/4} e^{3 \pi/4} (x+1)^{3/4}}{(1-x)^3} + \int_0^1 dx \frac{x^{1/4} e^{i 3 \pi/4} (1-x)^{3/4}}{(1+x)^3} \\ + \int_1^0 dx \frac{x^{1/4} e^{-i 3 \pi/4} (1-x)^{3/4}}{(1+x)^3} + e^{-i \pi} PV \int_0^R dx \frac{e^{-i \pi/4} x^{1/4} e^{-3 \pi/4} (x+1)^{3/4}}{(1-x)^3}\\ + \frac1{R} \int_{-\pi}^{\pi} d\theta \, e^{-i \theta} \left (1-\frac1{R e^{i \theta}} \right )^{3/4} \left (1+\frac1{R e^{i \theta}} \right )^{-3}\\ + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{i \pi/4}(1- \epsilon e^{i \phi})^{1/4} e^{i 3 \pi/4} (2-\epsilon e^{i \phi})^{3/4}}{\epsilon^3 e^{i 3 \phi}} \\ + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{e^{-i \pi/4}(1- \epsilon e^{i \phi})^{1/4} e^{-i 3 \pi/4} (2-\epsilon e^{i \phi})^{3/4}}{\epsilon^3 e^{i 3 \phi}}$$
The 1st and 4th integrals cancel.  We then take the limit as $R \to \infty$ and find that the fifth integral also vanishes.  (This means that the so-called "residue at infinity" vanishes.)  The 6th and 7th integrals (over the humps about the pole at $z=-1$) combine.  By Cauchy's theorem, the contour integral is zero.  Thus, we have
$$i 2 \sin{\frac{3 \pi}{4}}\int_0^1 dx \frac{x^{1/4} (1-x)^{3/4}}{(1+x)^3} = -\frac{i}{\epsilon^2} \int_0^{2 \pi} d\phi \, e^{-i 2 \phi} (1-\epsilon e^{i \phi})^{1/4} (2-\epsilon e^{i \phi})^{3/4} $$
Note that, as we expand out in powers of $\epsilon$, all contributions to the integral vanish except those of $O(\epsilon^2)$.  Expanding the integrand produces
$$\begin{align} 2^{3/4} \left (1-\epsilon e^{i \phi} \right )^{1/4} \left (1-\frac12 \epsilon e^{i \phi} \right )^{3/4}  &= 2^{3/4} \left (1-\frac14 \epsilon e^{i \phi} - \frac{3}{32} \epsilon^2 e^{i 2 \phi}+\cdots \right ) \left (1-\frac38 \epsilon e^{i \phi} - \frac{3}{128} \epsilon^2 e^{i 2 \phi}+\cdots \right ) \\ &= 2^{3/4} \left [1 + \frac18 \epsilon e^{i \phi} - \frac{3}{128} \epsilon^2 e^{i 2 \phi} + \cdots \right ] \end{align}$$
Thus,
$$\int_0^1 dx \frac{x^{1/4} (1-x)^{3/4}}{(1+x)^3} = 2 \pi \frac{2^{3/4} 3}{\sqrt{2} 128} = \frac{2^{1/4} 3 \pi}{64} $$
A: The solution based on beta function method provided by @EvilNebula is undoubtedly the most elegant route to go. For the sake of variety though, I thought it worth noting that the integrals in question do in fact reduce to integrals of rational functions, and thus are elementary. So the advanced beta function machinery, while time-saving, isn't strictly necessary here.

Consider the following generalization, which includes the two integrals from the OP as special cases:

For $m,n\in\mathbb{N}\land z\in\mathbb{R}\land0\le z<1$, define $I_{m,n}{\left(z\right)}$ via the integral representation
  $$I_{m,n}{\left(z\right)}:=\int_{0}^{z}\frac{\sqrt[n]{x\left(1-x\right)^{n-1}}}{\left(1+x\right)^{m}}\,\mathrm{d}x.$$

We find through substitutions that for all $m,n\in\mathbb{N}$ the integral $I_{m,n}{\left(z\right)}$ may be reduced to one with a rational integrand:
$$\begin{align}
I_{m,n}{\left(z\right)}
&=\int_{0}^{z}\frac{\sqrt[n]{x\left(1-x\right)^{n-1}}}{\left(1+x\right)^{m}}\,\mathrm{d}x\\
&=\int_{0}^{z}\frac{\left(1-x\right)\sqrt[n]{\frac{x}{1-x}}}{\left(1+x\right)^{m}}\,\mathrm{d}x\\
&=\int_{0}^{\frac{z}{1-z}}\left(\frac{1+y}{1+2y}\right)^{m}\frac{\sqrt[n]{y}}{1+y}\cdot\frac{\mathrm{d}y}{\left(1+y\right)^{2}};~~~\small{\left[\frac{x}{1-x}=y\right]}\\
&=\int_{0}^{\frac{z}{1-z}}\frac{\left(1+y\right)^{m-3}}{\left(1+2y\right)^{m}}\sqrt[n]{y}\,\mathrm{d}y\\
&=\int_{0}^{\sqrt[n]{\frac{z}{1-z}}}\frac{n\left(1+t^{n}\right)^{m-3}t^{n}}{\left(1+2t^{n}\right)^{m}}\,\mathrm{d}t;~~~\small{\left[\sqrt[n]{y}=t\right]}.\\
\end{align}$$
From there, we would apply partial fraction decomposition to reduce the integrand to a sum of reciprocals of powers linear and irreducible quadratic terms, and then integrate term by term in the usual manner to obtain a final value in terms of logarithms and inverse trigonometric functions.
