The integration is like: $$\int_{a}^{b}\left(\frac{b-x}{x-a}\right)^{p}dx$$ with $0<p<1$
Answer is $(b-a)p \frac{\pi}{\sin p\pi}$
Apparently, we can reversely construct $$\Gamma(1-p) \Gamma(p) = \frac{\pi}{\sin \pi p}$$ but when I continue I find that I can't transform the old interval into the $(0,1)$ which is needed by $B$ function.
Try to substitute $$x=(b-a)t+a \;\text{ where }t\in [0,1]$$ Then use the beta function $$B(x,y) =\int_{0}^{1} t^{x-1}(1-t)^{y-1} dt =\cfrac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$$