For example, consider:
$$I = \int_{z=0}^{d}\int_{r=0}^{a}\frac{zr}{\sqrt{z^2+r^2}} \ dr \ dz $$
Without the square root term, this would be a simple integral. With it, I can't find a way to solve it. Is there a general way to solve integrals with non-separable terms?
Double integrals which are "separable" are actually the exception - most are not. The general idea is that you evaluate a double integral as an integral of an integral. You also have to remember that in $\int\cdots dx$ there are implied brackets, so your integral needs to be considered as $$I = \int_{z=0}^{d}\left(\int_{r=0}^{a}\frac{zr}{\sqrt{z^2+r^2}} \ dr\right) \ dz\ .$$ Now you do the "inner" integral, treating $r$ as the variable and $z$ as a constant, so that it is not much different from $$\int_{r=0}^{a}\frac{5r}{\sqrt{5^2+r^2}} \ dr\ .$$ Once you have completed this integral, there will be no $r$ in the answer (because you have substituted $r=0$ and $r=a$) but there will still be a $z$. So to complete the job you have to integrate $$I=\int_{z=0}^{d} \hbox{previous answer}(z)\,dz\ .$$