How to understand this variable change in a double integral?

Question

I am solving the following problem: $$\int_{0}^{1}\int^{1}_{0}\frac{dx\cdot dy}{x^{-1}+|\ln(y)|-1}\leq 1.$$ After showing that $x^{-1}-1\geq |\ln(x)|$ for $x\in (0,1]$ we get that
$$\int_{0}^{1}\int^{1}_{0}\frac{dx\cdot dy}{x^{-1}+|\ln(y)|-1}\leq \int_{0}^{1}\int^{1}_{0}\frac{dx\cdot dy}{|\ln(x)|+|\ln(y)|} =\int_{0}^{1}\int^{1}_{0}\frac{dx\cdot dy}{|\ln(x\cdot y)|}.$$ Now in the solution I see the following variable change $y=u/x$ which should imply (according to the author) that
$$\int_{0}^{1}\int^{1}_{0}\frac{dx\cdot dy}{|\ln(x\cdot y)|} \color{red}{=} \int_{0}^{1}\left(\int^{1}_{u}\frac{dx}{x}\right)\frac{du}{|\ln(u)|}.$$ I am not sure how this happened. Could someone please explain this?

Answer 1

Since $x=u/y$, $u\le x\le 1$. Further, $dy=d(u/x)=du/x$, so the result follows.

Answer 2

We are going from $xy$ plane to $xu$ plane.

The rectangle $0\le x\le 1$, $0\le y\le 1$, under transformation $x=x$, $y=u/x$ transforms to the triangle $0\le x\le 1$, $0\le u\le x$ which is then described as $0\le u\le 1$, $0\le x\le u$

Note that the upper side of the rectangle $y=1$ maps to the upper side of the triangle namely $u=x$