$$ \int_0^\infty \frac{\cos xt}{1+t^2}\,\mathrm{d}t = \frac{\pi}{2}e^{-|x|}$$
To transform your integral into this form use $x \mapsto t \cdot a$. Now we can begin by taking the laplace transform of the integral
$$
\mathcal{L}(I) = \int_{0}^{\infty} \left( \int_{0}^{\infty} \frac{\cos(xt)}{1+t^2} \, \mathrm{d}t\right)e^{-sx}\,\mathrm{d}x
$$
The next step is to interchange the limits (fubinis theorem). Since
$|I|$, converges in the Riemann sense, so does $I$.
$$
\begin{align*}
\mathcal{L}(I) = \int_{0}^{\infty} \left(\int_{0}^{\infty} \frac{\cos(xt)}{1+t^2} e^{-sx}\,\mathrm{d}x \right)\mathrm{d}t
& = \int_{0}^{\infty} \frac{1}{1+t^2}\left(\int_{0}^{\infty} \cos(xt) e^{-sx}\,\mathrm{d}x \right)\mathrm{d}t \\
& = \int_{0}^{\infty} \frac{1}{1+t^2} \frac{s}{s^2+t^2} \mathrm{d}t
\end{align*}
$$
Where we used the laplace tranform of $\cos(\omega x)$ in the last transition.
The last expression can easilly be solved by partial fractions
$$
\frac{1}{1+t^2} \frac{s}{s^2+t^2} = \frac{s}{1-s^2} \left( \frac{1}{s^2+t^2} - \frac{1}{1+t^2} \right)
$$
To obtain the final answer you have to take the inverse-laplace transform of your integral.
\begin{align*}
|I|
\leq \int_{0}^{\infty} \int_{0}^{\infty} \frac{e^{-sx}}{1+t^2} \mathrm{d}t\,\mathrm{d}x
= \int_{0}^{\infty} \left[ \frac{\pi}{2}e^{-sx} \right]_0^\infty \mathrm{d}t\,\mathrm{d}x
= \biggl[ -\frac{\pi}{2s} e^{-sx} \biggr]_0^\infty
= \frac{\pi}{2s}
\end{align*}