Beta function-like integral $$\int\limits_0^1 \frac{x^{1-\alpha} (1-x)^\alpha}{(1+x)^3} \, dx $$
After the substitution $z=\frac{1}{x} - 1$,
I've got this:
$$\int\limits_0^\infty \left(\frac{1}{z}-1\right)^\alpha\left(\frac{1}{z}+1\right)^{-3} dz $$
But it is still far from the Beta function.


The key is the linear-fractional transformation,
$$x\mapsto\frac{1-t}{1+t}.$$
Then $1-x=\frac{2t}{1+t}$ and $1+x=\frac{2}{1+t}$, and we find
$$\begin{align}
\int_{0}^{1}\frac{x^{1-a}\left(1-x\right)^{a}}{\left(1+x\right)^{3}}\,\mathrm{d}x
&=\frac12\int_{1}^{0}\left(\frac{1-x}{x}\right)^{a}\frac{x}{1+x}\cdot\frac{\left(-2\right)}{\left(1+x\right)^{2}}\,\mathrm{d}x\\
&=\frac12\int_{0}^{1}\left(\frac{2t}{1-t}\right)^{a}\frac{1-t}{2}\,\mathrm{d}t;~~~\small{\left[x=\frac{1-t}{1+t}\right]}\\
&=\frac{1}{2^{2-a}}\int_{0}^{1}t^{a}\left(1-t\right)^{1-a}\,\mathrm{d}t\\
&=\frac{1}{2^{2-a}}\operatorname{B}{\left(1+a,2-a\right)}.\blacksquare\\
\end{align}$$
