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 substitution $u = \frac{1}{x}-1$ is actually useful because it transforms it into an integral that can be evaluated fairly easily using contour integration.
Specifically, for $-1 <a <2$,
$$\int_{0}^{1} \frac{x^{1-a}(1-x)^{a}}{(1+x)^{3}} \, dx = -\int^{0}_{\infty} \frac{\left(\frac{1}{1+u} \right)^{1-a} \left(1- \frac{1}{1+u} \right)^{a}}{\left(1+ \frac{1}{1+u} \right)^{3}} \frac{du}{(1+u)^{2}} = \int_{0}^{\infty} \frac{u^{a}}{(2+u)^{3}} \, du.$$
Integrating the function $$f(z) = \frac{z^{a}}{(2+z)^{3}}$$ around a keyhole contour deformed around a branch cut along the positive real axis, we get 
$$ \begin{align} \int_{0}^{\infty} \frac{u^{a}}{(2+u)^3} \, du + \int_{\infty}^{0} \frac{(ue^{2 \pi i})^{a}}{(2+u)^{3}} \, du &= 2 \pi i \, \operatorname{Res}[f(z), -2] \\ &= 2 \pi i \, \frac{1}{2!} \lim_{z \to -2} \frac{\mathrm{d}^{2}}{\mathrm{d}z^{2}} \, z^{a} \\ &=\pi i a(a-1)(-2)^{a-2} \\ &= \pi i a(a-1)e^{i \pi a}2^{a-2}. \end{align} $$
Therefore, $$\begin{align} \int_{0}^{\infty} \frac{u^{a}}{(2+u)^{3}} \, du &=  \pi 2^{a-2} a(a-1) \, \frac{i e^{\pi ia}}{1-e^{2 \pi ia}}\\  &=  \pi 2^{a-3} a(1-a) \csc(\pi a) . \end{align} $$ 
This is the result you'll get if you apply the functional equation and the reflection formula of the gamma function to David H's answer.
When $a=0$ or $a=1$, the result should be interpreted as a limit.
