Evaluating $\iint_{[0,1]^2} \frac{2-4xy}{(9-xy)(8+xy)}dxdy$ I am trying to compute the following double integral:
$$I=\iint_S \frac{2-4xy}{(9-xy)(8+xy)}dxdy$$
with $S=[0,1]\times[0,1].$
What I have tried:


*

*I have written the integral as follows:
$$I=I_1+I_2=-2\iint_S \frac{1}{(9-xy)}dxdy+2\iint_S \frac{1}{(8+xy)}dxdy$$
When trying to compute $I_1$, I encountered this integral $\int_0^1-\frac{1}{x}\ln(1-\frac{x}{9})dx$. I used an online calculator to solve it and the result involves the function $\operatorname{Li}(z)$, which I am not familiar with.

*Another thing I have tried is a change of variables:
$$u=9-xy$$ $$v=8+xy$$
The problem is that the Jacobian associated to this change of variables is null, so I cannot use it.


My question:
Could someone show me a way of computing this integral without having to use $\operatorname{Li}_2(z)$?
 A: In the following I'll address a solution in one line to the OP's question in comments. We may observe the simple facts that $\int_0^1 \frac{\log(t)}{a+1-t}\textrm{d}t=\int_0^1 \frac{1/(1+a)\log(t)}{1-(1/(1+a))t}\textrm{d}t=-\operatorname{Li}_2\left(\frac{1}{1+a}\right)$ and $\int_0^1 \frac{\log(t)}{a+t}\textrm{d}t=\int_0^1 \frac{1/a\log(t)}{1+(1/a) t}\textrm{d}t=\operatorname{Li}_2\left(-\frac{1}{a}\right)$ based on the integral representation of Polylogarithm, $\int_0^1\frac{u \log^n(x)}{1-u x}\textrm{d}x=(-1)^n n!\operatorname{Li}_{n+1}(u)$. A solution may be found in the book, (Almost) Impossible Integrals, Sums, and Series, pages $70-71$. 

Now, splitting the main integral and using the previous results, we get
  $$\int_0^1 \log(t)\left(\frac{1}{a+1-t}-\frac{1}{a+t}\right)\textrm{d}t=-\operatorname{Li}_2\left(-\frac{1}{a}\right)-\operatorname{Li}_2\left(\frac{1}{1+a}\right)=\frac{1}{2}\log^2\left(1+\frac{1}{a}\right),$$ 
  where the last equality comes from Landen's identity.

End of story.
A: Let's generalize it, in particular the integral in the question is just the case $a=8$.
$$I(a)=\int_0^1\int_0^1\left(\frac{1}{a+xy}-\frac{1}{1+a-xy}\right)dxdy\overset{xy=t}=\int_0^1\int_0^y\frac{1}{y}\left(\frac{1}{a+t}-\frac{1}{1+a-t}\right)dtdy$$
$$=\int_0^1\int_t^1\frac{1}{y}\left(\frac{1}{a+t}-\frac{1}{1+a-t}\right)dydt=\int_0^1\ln t \left(\frac{1}{1+a-t}-\frac{1}{a+t}\right)dt$$
$$\overset{\large \frac{1+a-t}{a+t}=x}=\int_{a/(1+a)}^{(1+a)/a}\ln\left(\frac{(1+a)-ax}{1+x}\right)\frac{1-x}{1+x}\frac{dx}{x}\overset{x\to 1/x}=\int_{a/(1+a)}^{(1+a)/a}\ln\left(\frac{1+x}{(1+a)x-a}\right)\frac{1-x}{1+x}\frac{dx}{x}$$
By averaging the two integrals from above we obtain:
$$I(a)=\frac12 \int_{a/(1+a)}^{(1+a)/a}\ln\left(\frac{(1+a)-ax}{(1+a)x-a}\right)\frac{1-x}{1+x}\frac{dx}{x}$$
Similarly, via the substitution $x\to \frac{1}{x}$ we can see that:
$$\int_{a/(1+a)}^{1}\ln\left(\frac{(1+a)-ax}{(1+a)x-a}\right)\frac{1-x}{1+x}\frac{dx}{x}=\int_{1}^{(1+a)/a}\ln\left(\frac{(1+a)-ax}{(1+a)x-a}\right)\frac{1-x}{1+x}\frac{dx}{x}$$
$$\Rightarrow I(a)=\int_{1}^{(1+a)/a}\ln\left(\frac{(1+a)/a-x}{(1+a)/a x-1}\right)\frac{1-x}{1+x}\frac{dx}{x}$$
$$\overset{\large \frac{(1+a)/a-x}{(1+a)/a x-1}=t}=\int_0^1 \ln t \left(\frac{2}{1+t}-\frac{1}{(1+a)/a+t}-\frac{1}{a/(1+a)+t}\right)dt$$
Now we will use the substitution $t=k x$, where $k$ is the constant found in each denominator.
$$I(a)=2\int_0^1 \frac{\ln x}{1+x}dx-\int_0^{(1+a)/a}\frac{\ln\left(\frac{1+a}{a}x\right)}{1+x}dx-\int_0^{a/(1+a)}\frac{\ln\left(\frac{a}{1+a}x\right)}{1+x}dx$$
$$=-\ln\left(\frac{1+a}{a}\right)\ln\left(1+\frac{1+a}{a}\right)-\ln\left(\frac{a}{1+a}\right)\ln\left(1+\frac{a}{1+a}\right)$$
$$+2\int_0^1 \frac{\ln x}{1+x}dx-\int_0^{(1+a)/a}\frac{\ln x}{1+x}dx-\int_0^{a/(1+a)}\frac{\ln x}{1+x}dx$$
$$=\ln^2\left(\frac{1+a}{a}\right)+\int_{(1+a)/a}^1 \frac{\ln x}{1+x}dx+\int_{a/(1+a)}^1 \frac{\ln x}{1+x}dx$$
Again, the substitution $x\to \frac{1}{x}$ yields:
$$\int_{a/(1+a)}^1 \frac{\ln x}{1+x}dx=\int_{(1+a)/a}^1 \frac{\ln x}{x}dx-\int_{(1+a)/a}^1 \frac{\ln x}{1+x}dx$$
$$\Rightarrow I(a)=\ln^2\left(\frac{1+a}{a}\right)+\int_{(1+a)/a}^1 \frac{\ln x}{x}dx=\boxed{\frac12 \ln^2\left(\frac{1+a}{a}\right)},\ a>0$$
