Why are these two integrals different?

Take $f(x) = \ln(x)$

Then

$\int_{[0,1]} \int_{[0,1]} f(xy)dxdy = \int_{[0,1]} \int_{[0,1]} \ln(xy) dx dy = \int_{[0,1]} \int_{[0,1]} ( \ln(x) + \ln(y)) dx dy = 2(x\ln(x) - x) |_{0}^{1} = -2 $

And $\int_{[0,1]} f(x^{2})dx = \int_{[0,1]} \ln(x^{2}) dx= \int_{[0,1]} 2\ln(x)dx = 2(x\ln(x) -x)|_{0}^{1} = -2$

The answer to Question 1 has to do with the fact that in the first case, $a_n$ and $b_n$ are unordered, so the joint distribution of $\{(a_r, b_r)\}_{r \ge 1}$ is uniform on the unit square $(0,1)^2$. But in the second expression both $A_r/A_n$ and $B_r/B_n$ are ordered, so their joint distribution is no longer uniform on the unit square. Informally speaking, the product $$\frac{A_r B_r}{A_n B_n}$$ never has a term where $A_r/A_n$ is "small" but $B_r/B_n$ is "large" for a given index $r$, or vice versa. Each is uniform on $(0,1)$ but together, they are highly correlated--in the limit, perfectly correlated.