Is there a nice way to reconcile the antiderivatives of $\frac{1}{x^2+a^2}$, $\frac{1}{x^2-a^2}$, and $\frac{1}{x^2}$ as $a\to0$? If $a$ is a (WLOG) positive real number, ignoring all constants of integration we have
$$
\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right)
$$
$$
\int \frac{1}{x^2-a^2}\,dx = \frac{1}{2a}\log \left(\frac{a-x}{a+x}\right)
$$
$$
\int \frac{1}{x^2}\,dx = \frac{-1}{x}
$$Is it the case that the first two functions should approach $-x^{-1}$ in the limit as $a\to0$, and if so why? I can't imagine why or how we'd be able to pass the limit inside the integral first.

Remember that antiderivatives are only defined up to constant. One can select a sequence of constants $C_a$ so that $F_a(x)+C_a\to-1/x$ when $a\to0$ and $x\neq0$ for both those functions. It is easier to see with definite integrals like $\int_1^x \frac{1}{t^2+a^2}\,dt$, where it is justified to move the limit inside the integral because the convergence on $[1,x]$ is uniform. This gives us for $a\to0$:
$$
\int_1^x \frac{1}{t^2+a^2}\,dt=\frac{1}{a}\arctan\left(\frac{x}{a}\right)-\frac{1}{a}\arctan\left(\frac{1}{a}\right)\\
\to\int_1^x \frac{1}{t^2}\,dt=-\frac{1}{x}+1.
$$
Therefore,
$$
\frac{1}{a}\arctan\left(\frac{x}{a}\right)-\left(\frac{1}{a}\arctan\left(\frac{1}{a}\right)+1\right)\to -\frac{1}{x}.
$$
The other case is analogous. There is a deeper sense to reconciling those two cases that comes from passing to the complex domain, namely:
$$
\arctan(x)=\frac{1}{2i}\ln \left(\frac{x-i}{x+i} \right)+C.
$$
