Let's compute the Fourier Transform of the surface measure on a sphere of radius $r$. We will do this by integrating in slices.
The singular measure supported on the sphere is the limit of the volume measure on a thin sphere of thickness $\mathrm{d}r$ divided by $\mathrm{d}r$. When integrating a slice, the angle of intersection of the slice with the sphere must be taken into account. In the diagram below, it is shown that a surface of thickness $\mathrm{d}r$ intersecting a surface of thickness $\mathrm{d}x$ at an angle of $\theta$ has a cross sectional area of $\mathrm{d}x\,\mathrm{d}r\sec(\theta)$. $\hspace{2cm}$
When integrating along the slice whose intersection with the sphere is a circle of radius $r\cos(\theta)$, the angle of intersection of the slice with the surface of the sphere is $\theta$. Thus, the $\sec(\theta)$ from the angle of the intersection is cancelled by the $\cos(\theta)$ from the radius of the circle. Therefore, $ \begin{align} \int_{rS^2}e^{-2\pi i\,x\cdot\xi}\,\mathrm{d}x &=\int_{-r}^r2\pi re^{-2\pi i\,t|\xi|}\mathrm{d}t\\ &=\frac{r}{-i|\xi|}\left(e^{-2\pi i\,r|\xi|}-e^{2\pi i\,r|\xi|}\right)\\ &=\frac{2r}{|\xi|}\sin(2\pi r|\xi|)\tag{1} \end{align} $ Computing the Fourier Transform
To compute the Fourier transform of $\dfrac{1}{r^2+1}$, we integrate against $(1)$: $ \begin{align} \int_0^\infty\frac{2r}{|\xi|}\frac{\sin(2\pi r|\xi|)}{r^2+1}\mathrm{d}r &=\int_{-\infty}^\infty\frac{r}{|\xi|}\frac{\sin(2\pi r|\xi|)}{r^2+1}\mathrm{d}r\\ &=\frac{1}{|\xi|}\int_{-\infty}^\infty\frac{r\sin(r)\,\mathrm{d}r}{r^2+4\pi^2|\xi|^2}\\ &=\frac{1}{|\xi|}\Im\left(\int_{-\infty}^\infty\frac{re^{ir}\,\mathrm{d}r}{r^2+4\pi^2|\xi|^2}\right)\\ &=\frac{1}{|\xi|}\Im\left(\int_\gamma\frac{re^{ir}\,\mathrm{d}r}{r^2+4\pi^2|\xi|^2}\right)\\ &=\frac{1}{|\xi|}\Im\left(2\pi i\operatorname{Res}\left(\frac{re^{ir}}{r^2+4\pi^2|\xi|^2},2\pi i|\xi|\right)\right)\\ &=\frac{1}{|\xi|}\Im\left(2\pi i\frac{2\pi i|\xi|e^{-2\pi|\xi|}}{4\pi i|\xi|}\right)\\ &=\frac{\pi}{|\xi|}e^{-2\pi|\xi|}\tag{2} \end{align} $ where $\gamma$ is the limit of the path on the real axis from $-M$ to $M$ followed by the semi-circle in the upper half-plane centered at $(0,0)$ from $M$ back to $-M$ as $M\to\infty$.
Therefore, $ \int_{\mathbb{R}^3}\frac{1}{|x|^2+1}e^{-2\pi i\,x\cdot\xi}\;\mathrm{d}x=\frac{\pi}{|\xi|}e^{-2\pi|\xi|}\tag{3} $
Relation to the Laplacian Equation
Taking the Fourier Transform of $ u(x)-\Delta u(x)=f(x)\tag{4} $ yields $ (1+4\pi^2|\xi|^2)\hat{u}(\xi)=\hat{f}(\xi)\tag{5} $ which becomes $ \hat{u}(\xi)=\dfrac{1}{1+4\pi^2|\xi|^2}\hat{f}(\xi)\tag{6} $ and taking the Inverse Fourier Transform yields $ \begin{align} u(x) &=\left(\dfrac{1}{1+4\pi^2|\xi|^2}\right)^\wedge(x)\ast f(x)\\ &=\frac{1}{4\pi|x|}e^{-|x|}\ast f(x)\tag{7} \end{align} $ The convolution in $(7)$ is a Singular Integral.