Differentiating under the integral sign:
Clearly the integral is just
$$ \DeclareMathOperator{\diff}{\,d\!}
2 \int_0^{+\infty} \frac {\exp(-ax^2)}{x^2+b^2} \diff x.
$$
Let the integral be $I(a)$ and we regard $b$ as a constant. We want to differentiate w.r.t. $a$. For each $c > 0$, on $J_c=[c, +\infty)$, the integral converges for $a \in J_c$, and
$$\newcommand{\Abs}[1]{{\left|#1\right|}}
\Abs{\partial_a \frac {\exp(-ax^2)}{x^2+b^2}} \leqslant \exp(-cx^2),
$$
where $\int_0^{+\infty} \exp(-cx^2) \diff x$ converges, so by Weierstrass M-test,
$$
\int_0^{+\infty} \frac {\exp(-ax^2)(-x^2)}{x^2+b^2}\diff x
$$
converges uniformly for $a\in J_c$. Therefore $I(a)$ could be differentiated under the integral symbol. Now
\begin{align*}
I'(a) &= \int_0^{+\infty} \frac {\exp(-ax^2) (-x^2)}{x^2+b^2}\diff x\\
&= \int_0^{+\infty} \exp(-ax^2) \left( \frac {b^2}{x^2+b^2}-1 \right)\diff x\\
&= b^2 I(a) - \int_0^{+\infty} \exp(-ax^2)\diff x\\
&= b^2 I(a) - \frac {\sqrt \pi} {2\sqrt a}.
\end{align*}
Now solve this differential equations under the initial value condition
$$
I(0) = \int_0^{+\infty}\frac {\diff x}{x^2+b^2} = \frac \pi{2b}.
$$
By the formula for 1st-order ODE, we have
$$
I(a) = \frac{\exp(b^2a)}b \left(-\int_0^{b\sqrt a} \exp(-u^2)\diff u + \frac \pi {2b}\right),
$$
and the original is just $2I(a)$.