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.

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
$$\Large\color{blue}{I\left ( 1,4 \right )=\frac{3\sqrt[4]{2}}{64}\pi }$$
$$\Large\color{blue}{I\left ( 1,3\right )=\frac{\pi }{18}\frac{\sqrt[3]{4}}{\sqrt{3}}}$$
