What is the rule saying that $\int_{-2}^2 \frac{1}{x^2-4}\mathrm dx$ diverges? Problem
Calculate $\int\limits_{-2}^2 \frac{1}{x^2-4}\mathrm dx$ unless it diverges.
My attempt
I calculated the indefinite integral $$\int\frac{1}{x^2-4}\mathrm dx = \frac14\left( \ln(2-x) - \ln(2+x) \right) + C$$
and notice from there that the limits on the integral will cause $\ln(0)$ problems.
Question
From this I decided that the integral must be divergent, but does that hold? Or is there some simple rule that would have allowed me to conclude this before calculating the indefinite integral?
Any tips appreciated!
 A: Heuristically, you could have noted that $x^2-4$ has nonzero, finite slope near $x=2$, so it behaves approximately like $c(x-2)$ for some $c$ there.
Therefore its reciprocal will diverge mostly in the same way as $\frac{1}{c(x-2)}$, and we know that the integral of that kind of diverging functions will itself diverge. (This would be in contrast to, for example, $c(2-x)^{-1/2}$, where the function itself diverges but the area below it is finite).
This could be made into a more rigorous proof by comparing it the integral of $\frac{1}{x^2-4}$ near $x=2$ with that of $\frac{1}{k(x-2)}$ for a sufficiently large $k$ that $|k(x-2)|>|x^2-4|$ in a punctured neighborhood of $2$. In this particular case that is probably not much of an improvement over finding the full indefinite integral, but it is valuable to be able to use the heuristic reasoning.
A: $\displaystyle \int\limits_{-a}^a\frac{dx}{x^2-4} = 2\int\limits_0^a\frac{dx}{x^2-4}=\frac{1}{4}\ln\frac{2-a}{2+a}$ with $0<a<2$ . 
Then we see what happens for $a\to 2^-$ .  
