$\int_{-1}^{1}\int_{-1}^{1}\frac{dxdy}{\sqrt{(x-.3)^2+y^2}}$ has the following physical meaning: it is the potential of the uniform surface source distributed on a square $|x|,|y|<1$ observed at a point $O=(0.3,0)$. Obviously the point $O$ is a singular point.
I could evaluate the following integrals using WOLFARM ALPHA: $$\int_{-1}^{1}\int_{-1}^{0.3}\frac{dxdy}{\sqrt{(x-.3)^2+y^2}}=3.9989$$ $$\int_{-1}^{1}\int_{0.3}^{1}\frac{dxdy}{\sqrt{(x-.3)^2+y^2}}=2.9216$$ The question is is it allowable to add up the results obove to conclude the following in spite of the above mentioned singularity? $$\int_{-1}^{1}\int_{-1}^{1}\frac{dxdy}{\sqrt{(x-.3)^2+y^2}}=3.9989+2.9216=6.9205$$ Any other way to solve?