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.
 A: 
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
$$\begin{align}
\int_{0}^{1}\frac{x^{1-a}\left(1-x\right)^{a}}{\left(1+x\right)^{3}}\,\mathrm{d}x
&=\frac12\int_{1}^{0}\left(\frac{1-x}{x}\right)^{a}\frac{x}{1+x}\cdot\frac{\left(-2\right)}{\left(1+x\right)^{2}}\,\mathrm{d}x\\
&=\frac12\int_{0}^{1}\left(\frac{2t}{1-t}\right)^{a}\frac{1-t}{2}\,\mathrm{d}t;~~~\small{\left[x=\frac{1-t}{1+t}\right]}\\
&=\frac{1}{2^{2-a}}\int_{0}^{1}t^{a}\left(1-t\right)^{1-a}\,\mathrm{d}t\\
&=\frac{1}{2^{2-a}}\operatorname{B}{\left(1+a,2-a\right)}.\blacksquare\\
\end{align}$$
A: The answer is 
$$\frac{\Gamma(2-\alpha)\Gamma(1+\alpha)}{\Gamma(2\alpha+3)}\ _{2}F_{1}(3, 2-\alpha; 2\alpha+3, -1)$$
Where $_{2}F_{1}(a, b; c; z)$ is the Gauss Hypergeometric function. See http://mathworld.wolfram.com/HypergeometricFunction.html
A: 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.
