Just basically independently reinvented Bryan Yock's solution as a more 'pure' version of Feynman.
Let $$I(b) = \int_0^\infty \frac {e^{-x^2}}{1+(x/b)^2} \mathrm d x = \int_0^\infty \frac{e^{-b^2y^2}}{1+y^2} b\,\mathrm dy$$ so that $I(0)=0$, $I'(0)= \pi/2$ and $I(\infty)$ is the thing we want to evaluate.
Now note that rather than differentiating directly, it's convenient to multiply by some stuff first to save ourselves some trouble. Specifically, note
$$\left(\frac 1 b e^{-b^2}I\right)' = -2b \int_0^\infty e^{-b^2(1+y^2)} \mathrm d y = -2 e^{-b^2} I(\infty)$$
Then usually at this point we would solve the differential equation for all $b$, and use the known information at the origin to infer the information at infinity. Not so easy here because the indefinite integral of $e^{-x^2}$ isn't known. But we don't actually need the solution in between; we only need to relate information at the origin and infinity. Therefore, we can connect these points by simply integrating the equation definitely; applying $\int_0^\infty \mathrm d b$ we obtain
$$-I'(0)= -2 I(\infty)^2 \quad \implies \quad I(\infty) = \frac{\sqrt \pi} 2$$
A direct application might be
$$\left.\frac{d}{ds} \int_0^{1-\delta} \frac{x^s}{1-x} \, dx\right|_{s = 0} = \left.\int_0^{1-\delta} \frac{ x^s \ln x}{1-x} \, dx\right|_{s = 0} = \int_0^{1-\delta} \frac{\ln x}{1-x} \, dx = \int_\delta^{1 } \frac{\ln (1-x)}{x} \, dx $$
Evaluate the integral on the LHS using the geometric series expansion expansion of $1/(1-x)$ and then take the limit as $\delta \to 0$ (since the improper integral on the RHS converges).
You can also proceed by proving the second equality in
$$- \int_0^1 \frac{\ln(1-x)}{x} \,dx = \int_0^1 \int_0^1 \frac{1}{1 - xy}\, dx \,dy = \sum_{k=1}^\infty \frac{1}{k^2} =\zeta(2) = \frac{\pi^2}{6}$$
using the geometric series $1/(1 -xy) = 1 + xy + (xy)^2 + \ldots $ The first equality is fairly obvious.
More generally we get by the same process
$$\int_0^1 \int_0^1 \frac{x^\alpha y^\alpha}{1 - xy}\, dx \,dy = \sum_{k=1}^\infty \frac{1}{(k + \alpha)^2}$$
and Feynman's trick of repeated integration with respect to $\alpha$ is used to extend the result to other integrals.
Best Answer
I'm not sure about Feynman's trick, but there is a delightfully simple way to compute this integral using Glasser's Master Theorem. First, substitute $u=x^2$ to obtain $$2\int_0^\infty \frac{u^2}{u^4+2u^2+5}\,du=\int_{-\infty}^\infty \frac{1}{u^2+5u^{-2}+2}\,du$$ Now, notice that $$u^2+5u^{-2}+2=\left(u-\frac{\sqrt{5}}{u}\right)^2+4\phi,$$ where $\phi=\frac{1+\sqrt{5}}{2}$. Glasser's Master Theorem then tells us that $$\int_{-\infty}^\infty \frac{1}{\left(u-\sqrt{5}/u\right)^2+4\phi}\,du=\int_{-\infty}^\infty \frac{1}{u^2+4\phi}\,du.$$ The last integral is just $$\frac{\pi}{2\sqrt{\phi}}=\frac{\pi}{\sqrt{2}}\cdot\frac{1}{\sqrt{1+\sqrt{5}}}$$