# Integral of (1+t)^(2x-1)(1-t)^(2y-1)(1+t^2)^(-x-y)dt=2^(x+y-2)B(x,y)

The following formula holds for $\mathrm{Re}(x)>0$ and $\mathrm{Re}(y)>0$: $$\displaystyle\int_{-1}^1 (1+t)^{2x-1}(1-t)^{2y-1}(1+t^2)^{-x-y} \mathrm{d}t= 2^{x+y-2}B(x,y),$$ where $B$ denotes the beta function.