= Solution
A <phase function> is a real smooth function
$$
\Phi:X\times(\mathbb R^k\setminus\{0\})\to\mathbb R
$$
that is <positively homogeneous function>[positively homogeneous] of degree one in $\theta$ and has no critical point in all variables:
$$
\Phi(x,t\theta)=t\Phi(x,\theta),
\qquad
\nabla_{x,\theta}\Phi(x,\theta)\ne0.
$$
The <symbol class> $\operatorname{Sym}(X,\mathbb R^k;N)$ consists of all $a\in C^\infty(X\times\mathbb R^k)$ such that, for every compact $K\Subset X$ and <multi-index>[multi-indices] $\alpha,\beta$,
$$
|D_x^\alpha D_\theta^\beta a(x,\theta)|
\leq C_{K,\alpha,\beta}(1+|\theta|)^{N-|\beta|}.
$$
For a cutoff $\chi\in C_c^\infty(\mathbb R^k)$ equal to one near zero, define the <oscillatory integral> by
$$
\boxed{
\langle I_\Phi(a),\psi\rangle
=\lim_{\varepsilon\downarrow0}
\int_X\int_{\mathbb R^k}
e^{i\Phi(x,\theta)}a(x,\theta)\psi(x)\chi(\varepsilon\theta)
\,d\theta\,dx}.
$$
Repeated <integration by parts> makes the limit meaningful and independent of the cutoff.
The <singular support> is the complement of the largest open set on which the distribution is represented by a <smooth function>. Suppose $x_0$ does not belong to
$$
C_\Phi=\{x:\nabla_\theta\Phi(x,\theta)=0
\text{ for some }\theta\ne0\}.
$$
On a sufficiently small neighborhood of $x_0$, homogeneity and compactness of the unit sphere give a lower bound for $|\nabla_\theta\Phi|$. The differential operator
$$
L=\frac{\nabla_\theta\Phi\mathbin\cdot\nabla_\theta}
{i|\nabla_\theta\Phi|^2}
$$
satisfies $Le^{i\Phi}=e^{i\Phi}$. Repeatedly transferring $L$ to the amplitude lowers its symbol order until the integral and all its $x$-derivatives converge absolutely. Hence $I_\Phi(a)$ is smooth near $x_0$, proving
$$
\boxed{\operatorname{sing\,supp}I_\Phi(a)\subseteq C_\Phi}.
$$
For the stated distribution on $\mathbb R^3$, rotational symmetry lets us align the polar axis with $x$ and write $\rho=|x|$. The angular integral is
$$
\int_0^{2\pi}\int_0^\pi
e^{ir\rho\cos\theta}\sin\theta\,d\theta\,d\phi
=4\pi\frac{\sin(r\rho)}{r\rho}.
$$
Therefore, for $\rho>0$,
$$
\begin{aligned}
u(x)
&=\frac1{2\pi}\frac{4\pi}{\rho}
\int_0^\infty\frac{r\sin(r\rho)}{1+r^2}\,dr\\
&=\boxed{\frac{\pi e^{-\rho}}{\rho}}.
\end{aligned}
$$
The original amplitude is not <Lebesgue integrable function>[Lebesgue integrable] in $k$, so this computation illustrates how oscillation assigns a distribution to a divergent ordinary integral. The result is smooth away from the origin and has singular support $\{0\}$. It is a constant multiple of the three-dimensional <Yukawa potential>, satisfying $(-\Delta+1)u=4\pi^2\delta_0$ with the normalization used in the question.
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