\(\iint_R\frac{1}{(x+y)^2}\,\dd(x,y)={}\) \(\ln(2)\)

It should be clear that the function \((x,y)\mapsto \frac{1}{(x+y)^2}\) is not defined at the point \((0,0)\) on the edge of the region of integration \(R\) and takes larger values as one approaches the origin. Yet the double integral has a finite value: \[\begin{aligned} \iint_R\frac{1}{(x+y)^2}\,\dd(x,y) &= \lim_{\xi\downarrow 0}\int_{x=\xi}^{x=1}\left(\int_{y=0}^{y=x^2}\frac{1}{(x+y)^2}\,\dd y\right)\dd x \\[0.25cm] &= \lim_{\xi\downarrow 0}\int_{x=\xi}^{x=1}\biggl[-\frac{1}{x+y}\biggr]_{y=0}^{y=x^2}\,\dd x\\[0.25cm] &= \lim_{\xi\downarrow 0}\int_{x=\xi}^{x=1}\left(\frac{1}{x}-\frac{1}{x^2+x}\right)\\[0.25cm] &=\int_{x=\xi}^{x=1} \frac{1}{x+1}\,\dd x\\[0.25cm] &=\int_{0}^{1} \frac{1}{x+1}\,\dd x\\[0.25cm] &= \biggl[\ln(x+1)\biggr]_{0}^{1}\\[0.25cm] &= \ln(2)-\ln(1)\\[0.25cm] &= \ln(2) \end{aligned}\]
We introduced a limit very formally here, but if you had calculated the iterated integral \(\displaystyle \int_{x=0}^{x=1}\left(\int_{y=0}^{y=x^2}\frac{1}{(x+y)^2}\,\dd y\right)\dd x\), then everything would have worked just fine and no improper integral would have appeared in the intermediate results.