This relates to my recent question:
Solving $\frac{\ln(x)\ln(y)}{\ln(1-x)\ln(1-y)}=1$ for $y$.
Now I'm trying to integrate the following expression but am having trouble putting it in a form in which I can integrate it.
$$\frac{\ln(x)\ln(y)}{\ln(1-x)\ln(1-y)}=n$$
for $n\in\Re(0,\infty)$ and $\ln(1-x)\ln(1-y) \ne0.$ For $n=1$ it's obvious that the area is $1/2.$ However I can't be exact for other values of $n.$
I tried applying the knowledge I learned from the answer to my recent question, but have not had any luck so far.