= Solution
For fixed $u\in(0,1)$, $x=1-u\in(0,1)$ and
$$
\frac{\partial y}{\partial v}=-\frac{1-u}{(1-uv)^2}<0.
$$
As $v$ 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 inverse
$$
u=1-x,\qquad v=\frac{1-y}{1-(1-x)y}.
$$
Its denominator is positive in the open square. This is a <rational square diffeomorphism>. Its <Jacobian determinant> is
$$
\frac{\partial(x,y)}{\partial(u,v)}
=(-1)\left(-\frac{1-u}{(1-uv)^2}\right)
=\frac{x}{(1-uv)^2}.
$$
Writing $A=1-(1-x)y$, the inverse relation gives $1-uv=x/A$. Consequently
$$
\boxed{\frac{\partial(x,y)}{\partial(u,v)}=\frac{A^2}{x}.}
$$
The 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> is
$$
\frac{\partial(u,v)}{\partial(t,w)}=\frac{1-w}{(1-wt)^2}.
$$
To 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 gives
$$
w=\frac{xy}{A},\qquad t=\frac{xA}{B},\qquad
1-w=\frac{1-y}{A},\qquad1-wt=\frac{1-y}{B}.
$$
The <chain rule> for <Jacobian determinants> therefore gives
$$
\boxed{\frac{\partial(x,y)}{\partial(t,w)}
=\frac{A^2}{x}\frac{B^2}{A(1-y)}
=\frac{AB^2}{x(1-y)}.}
$$
The requested integrand is exactly the reciprocal of this positive determinant. Apply the <change of variables formula> using the composite <diffeomorphism>:
$$
\boxed{\int_R\frac{x(1-y)}{[1-(1-x)y][1-(1-x^2)y]^2}\,dx\,dy
=\int_0^1\int_0^1 1\,dt\,dw=1.}
$$
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.
Back to article page
