<s>[INST] Computing $\int_{-a}^a \int_{-b}^b \frac 1{(x^2 + y^2 +c^2)^{3/2}}\, dxdy$. The question is exactly as in the title: 

$$\int_{-a}^a \int_{-b}^b \frac 1{(x^2 + y^2 +c^2)^{3/2}}\, dxdy$$

It's been so long since the last time I tried to calculate something like this. I first thought about polar coordinates but that doesn't go well with the domain of integration.
What kind of substitution do we need for this kind of problem? I am sorry if similar problem has been asked, I just couldn't manage to find it.
 [/INST] The inner integral is simple even without any substitution because ot the exponent $\frac 32$ in the denominator. Just write
$$\int \frac {dx}{(x^2 + y^2 +c^2)^{3/2}}=\frac {P(x)}{(x^2 + y^2 +c^2)^{1/2}}$$  Differentiate both sides, simplify and identify to get
$$\left(c^2+x^2+y^2\right) P'(x)-x P(x)=1$$ which is separable and the solution is
$$P(x)=\frac{x}{(c^2+y^2)}+C_1 \sqrt{c^2+x^2+y^2}$$ So
$$\int \frac {dx}{(x^2 + y^2 +c^2)^{3/2}}=\frac{1}{(c^2+y^2)}\frac {x}{(x^2 + y^2 +c^2)^{1/2}}$$
$$\int_{-b}^b \frac {dx}{(x^2 + y^2 +c^2)^{3/2}}=\frac{2 b}{\left(c^2+y^2\right) \sqrt{b^2+(c^2+y^2)}}$$ Now, as J.G. commented, a trigonometric change of variable would make the problem quite simple for the outer integral.
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