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For fixed $u\in (0,1)$, $x=1-u\in (0,1)$ andAs $v$ runs from zero to one, $y$ decreases continuously from one to zero. Thus the map is onto the open unit square and is one-to-one, with smooth inverseIts denominator is positive in the open square. This is a rational square diffeomorphism. Its Jacobian determinant isWriting $A=1-(1-x)y$, the inverse relation gives $1-uv=x/A$. ConsequentlyThe second transformation also maps the open unit square bijectively onto the open unit square: for fixed $w$, $u=(1-t)/(1-wt)$ decreases strictly from one to zero as $t$ goes from zero to one, while $v=1-w$ independently spans $(0,1)$. Its positive Jacobian determinant isTo express this in $x,y$, the composite transformation gives $y=w/[x+(1-x)w]$ and $x=t(1-w)/(1-wt)$. Put $B=1-(1-x^2)y$. Solving these relations givesThe chain rule for Jacobian determinants therefore givesThe requested integrand is exactly the reciprocal of this positive determinant. Apply the change of variables formula using the composite diffeomorphism:Possible singular behavior at the square's boundary does not invalidate the calculation: first integrate over the images of compact interior squares, where all transformations are smooth with nonzero Jacobian determinant, and then increase these domains to the full square. Positivity and monotone convergence theorem justify this limit.