Integral over a Really Weird Region

This picture shows roughly the area we want to integrate in(I am very very poor at drawing)

We have to use a variable change to solve the problem

Let $\displaystyle \frac{x^{2} + y^{2}}{x} = u$ and $\displaystyle \frac{x^{2}+y^{2}}{y} = v$

$\displaystyle \left\lvert \frac{\partial(u,v)}{\partial(x,y)} \right\rvert = \left(\frac{x^{2}+y^{2}}{xy}\right)^{2}$

So $\displaystyle \left\lvert \frac{\partial(x,y)}{\partial(u,v)} \right\rvert = \left(\frac{xy}{x^{2}+y^{2}}\right)^{2}$

So after this our integral just becomes a really simple one:

$\displaystyle \iint_{E} \frac{1}{xy} dx\,dy = \iint_{T} \frac{1}{xy} \left(\frac{xy}{x^{2}+y^{2}}\right)^{2} du\,dv =\iint_{T}\frac{xy}{(x^{2}+y^{2})^{2}}du\,dv$

Where $T$ is the rectangular region in $uv$ plane given by $1\leq u \leq 2$ and $1\leq v\leq 2$

So we have :- $\displaystyle \int_{1}^{2}\int_{1}^{2} \frac{1}{uv}\, du\,dv$

So $\displaystyle A = (\ln(2))^{2}$

$\lfloor{1000A}\rfloor = 480$

Monte Carlo with Python.

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import  random
n=10000000
f=0
for k in range(n):
  x=random.uniform(0.25,1)
  y=random.uniform(0.25,1)
  if x*x+y*y>x and  x*x+y*y<=2*x and  x*x+y*y<=2*y and   x*x+y*y>y:
    f=f+1/(x*y)
print(f/n*(9/16)*1000) 

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480.29366431811883