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?