Suppose I am evaluating the integral
$$\int_0^1\int_0^1\frac{y}{1+xy}dxdy$$
So I consider the following indefinite integral
$$\int\frac{y}{1+xy}dx=y\cdot\frac{1}{y}\int\frac{1}{1+xy}d(1+xy)=\ln|1+xy|$$
But since it may happen that $y=0$, I am wondering if I can do so legally.
Regardless of the intermediate calculations, the fact that $\frac{\partial}{\partial x}\left(\log|1+xy|\right) = \frac{y}{1+xy}$, immediately proves that, in the domain of definition of both expressions ($1+xy \ne 0$), you have $\int \frac{y}{1+xy} dx = \log|1+xy| + C$. So, regarding the base problem, you can write that $$ \int_0^1 \int_0^1 \frac{y}{1+xy} dx dy = \int_0^1(\log(1+y) - \log 1) dy = [(1+y)\log(1+y)-y]_0^1 = \log 4 -1 $$