Show that $\int_0^2 \int_0^2 \frac{x}{1+\ln(x^2y^2)} \,\mathrm{d}x \,\mathrm{d}y \leq 4$ 
Show that
  $$
  \int_0^2 \int_0^2 \frac{x}{1+\ln(x^2y^2)} \,\mathrm{d}x \,\mathrm{d}y \leq 4.
$$

When I think somewhat outside the box, I can see that when I sketch the boundaries in a one-dimensional way, it is clearly a square area with a side length of $2$, and obviously $2^2 = 4$. So in theory, if the definite integral is finding the area in certain contraints, the area has to be $\leq 4$ 
Is this the correct approach to this kind of question? And if so where should I go next?
This is only worth $2$ of $25$ marks which would indicate to me that I am not expected to actually complete the integration.
Any help is much appreciated. 
 A: About thinking outside the box: one of the main tricks in Math is to exploit the exchange of sums/limits/integrals. This trick has many forms: double counting, Feynman's trick, Wilf-Zeilberger couples, Fubini's theorem, creative telescoping... they all boil down to the same powerful idea. In our case we may try to write $\frac{1}{1+2\log(xy)}$ as $\int_{0}^{+\infty}e^{-t} x^{-2t} y^{-2t}\,dt $ to get...
$$ \iint_{(1,2)^2}\frac{x}{1+2\log(xy)}\,dx\,dy = \int_{0}^{+\infty} e^{-t}\iint_{(1,2)^2} x^{1-2t} y^{-2t}\,dx\,dy\,dt $$
(the situation gets temporarily worse) then
$$ \iint_{(1,2)^2}\frac{x}{1+2\log(xy)}\,dx\,dy = \int_{0}^{+\infty}\frac{(4^t-2)(4^t-4)}{(2t-2)(2t-1)} 4^{-2t} e^{-t}\,dt $$
(which is much better, since we have an integral in a single variable). Despite its algebraic nastyness the function $\frac{(4^t-2)(4^t-4)}{(2t-2)(2t-1)} 4^{-2t}$ has a very simple behaviour on $\mathbb{R}^+$: it is positive, decreasing and bounded above by $\frac{3}{2}\,\exp\left(-\frac{7}{5}t\right)$. In particular the original integral over $(1,2)^2$ does not exceed $\frac{5}{8}$. I leave to you to adapt this approach for dealing with the contribution over $(0,1)\times(1,2)$, $(1,2)\times(0,1)$ and $(0,1)^2$. 
$4$ is actually a very loose upper bound for the original integral: a sharper bound is $I\leq \frac{17}{14}$.
A: Ok so for anyone interested in the required answer, I've been told that for my level of mathematics, we can simply state: because $\int_0^2\int_0^2xdxdy=4$, and as $x$ divided by anything greater than one, as stipulated in the inequality, will always be $\le{x}$, the double integral will be$\int_0^2\int_0^2{x\over1+log(1 + x^2y^2)}dxdy\le4$
