If If Im,n=∫01xm-1(1-x)n-1dx:m,n≥1 and ∫01xm-1+xn-1(1+x)m+ndx=αIm,n,α∈R, then α=

Given Im,n=∫01xm-1(1-x)n-1dx Suppose that x=sin2θ,dx=2sinθcosθ The given integral becomes Im,n=2∫π2sin2m-1θcos2n-1θdθαIm,n=∫01xm−1+xn−11+xm+ndx                             ·x=tan2θ,dx=2tanθsec2θ=2∫0π4tan2m−1θ+tan2n−1θsec2m+2n−2θdθ=2∫0π2sin2m−1θcos2n−1θdθ=Im,nTherefore, α=1