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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.

John Zimmerman
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    What is $\mathfrak R(0,\infty)$? Do you mean the set of positive reals? – YiFan Tey Dec 23 '19 at 23:57
  • Yep that’s exactly what I mean yifan – John Zimmerman Dec 24 '19 at 00:02
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    I've never seen that denoted that way before; its usually just $(0,\infty)$, or some variant of $\mathbb R_{>0}$ with \mathbb{R} instead of \mathfrak{R}. – YiFan Tey Dec 24 '19 at 00:04
  • Or if you are lazy like me, just \Bbb R – Paul Sinclair Dec 24 '19 at 13:56
  • The vast majority of continuous functions you can write out do not have easily expressed anti-derivatives. When you cannot even find a closed expression for $y$ in terms of $x$, you are not likely to find a nice anti-derivative here. If it is even possible to find a simple expression for the integral at all, it will probably require other techniques. – Paul Sinclair Dec 24 '19 at 14:11

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