Simplify convolution integral $\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt$ I am trying to show that $$\frac{1}{a^2+x^2} \ast \frac{1}{a^2+x^2}=\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt = \frac{2\pi}{a(4a^2+x^2)}$$  I wrote out the integral for the convolution and it becomes a big exercise in partial fractions and integration of rational functions.
$$ \int(\frac{1}{a^2+t^2})  (\frac{1}{a^2+(x-t)^2}) dt=
\int\frac{x + 2 t}{x (4 a^2 + x^2) (a^2 + t^2)} + \frac{3 x - 2 t}{x (4 a^2 + x^2) (a^2 + x^2 - 2 xt + t^2)} dt =$$
$$=\frac{a (log(a^2 + t^2) - log(a^2 + (x - t)^2)) + x \cdot tan^{-1}(\frac{t}{a}) + x \cdot tan^{-1}(\frac{t - x}{a})}
{a x (4 a^2 + x^2)}
$$
Plugging in the bounds of plus/minus infinity makes the logs cancel out, and gives the 2$\pi x$ from the arctan terms.
My question: Is there a way to simplify / make this any easier using the properties of the convolution?  Or are these problems just 'plug and chug' like back in my engineering days?

Here using Fourier transform may simplify things considerably:
Let $\phi(x)=\frac{1}{1+x^2}$, and $\phi_a(x)=\frac{1}{a}\phi(a^{-1}x)=\frac{a}{a^2+x^2}$.
Recall that $\pi e^{-2\pi|t|}=\int e^{-2\pi itx}\frac{1}{1+x^2}\,dx=\widehat{\phi_1}(t)$. Hence
$$\widehat{\phi_a}(t)=\frac{1}{a}\int e^{-2\pi itx}\phi(xa^{-1})\,dx=\frac{1}{a}\int e^{2\pi ia t(a^{-1}x)}\phi(a^{-1}x)\,dx=\widehat{\phi_1}(at)$$
Thus
$$\widehat{\phi_a*\phi_a}(t)=(\widehat{\phi_a}(t))^2=(\widehat{\phi_1}(at))^2=\pi^2e^{-4\pi a|t|}$$
That is,
$$
I_a=a^{-2}(\phi_a*\phi_a)(x)=a^{-2}\pi^2\mathcal{F}^{-1}(e^{-4\pi a|t|})(x)=\frac{2\pi}{a(a^2+x^2)}$$
