Consider the transformation of variables

$$x=1-u, \quad y=\frac{1-v}{1-u v} .$$

Show that the interior of the unit square in the $u v$ plane

$$\{(u, v): 0<u<1,0<v<1\}$$

is mapped to the interior of the unit square in the $x y$ plane,

$$R=\{(x, y): 0<x<1,0<y<1\} .$$

[Hint: Consider the relation between $v$ and $y$ when $u=\alpha$, for $0<\alpha<1$ constant.]

Show that

$$\frac{\partial(x, y)}{\partial(u, v)}=\frac{(1-(1-x) y)^{2}}{x}$$

Now let

$$u=\frac{1-t}{1-w t}, \quad v=1-w$$

By calculating

$$\frac{\partial(x, y)}{\partial(t, w)}=\frac{\partial(x, y)}{\partial(u, v)} \frac{\partial(u, v)}{\partial(t, w)}$$

as a function of $x$ and $y$, or otherwise, show that

$$\int_{R} \frac{x(1-y)}{(1-(1-x) y)\left(1-\left(1-x^{2}\right) y\right)^{2}} d x d y=1$$