= Solution
Differentiate under the <integral> and use $\nabla_x|x-x'|^{-1}=-\nabla_{x'}|x-x'|^{-1}$:
$$
\begin{aligned}
\nabla\cdot A
&=-\frac{\mu_0}{4\pi}\int J_i(x')
\partial_i'\frac1{|x-x'|}\,d^3x'\\
&=\frac{\mu_0}{4\pi}\int
\frac{\nabla'\cdot J(x')}{|x-x'|}\,d^3x'.
\end{aligned}
$$
The omitted boundary term vanishes if the current is localized sufficiently rapidly. A steady current obeys charge conservation $\nabla'\cdot J=0$, hence
$$
\boxed{\nabla\cdot A=0}.
$$
For $r=|x|$ much larger than the source size,
$$
\frac1{|x-x'|}=\frac1r+\frac{x\cdot x'}{r^3}+O(r^{-3}|x'|^2).
$$
Localization and $\nabla'\cdot J=0$ imply $\int J\,d^3x'=0$. They also imply
$$
\int(x_i'J_j+x_j'J_i)\,d^3x'=0,
$$
by integrating $\partial_k'(x_i'x_j'J_k)$. Thus the first nonzero moment is antisymmetric and can be written using the <magnetic dipole moment>
$$
\boxed{m=\frac12\int x'\times J(x')\,d^3x'}.
$$
Consequently the <Coulomb-gauge vector potential of a localized steady current> has far field
$$
\boxed{A(x)=\frac{\mu_0}{4\pi}\frac{m\times x}{r^3}+\cdots}.
$$
The dimensions are
$$
\boxed{[J]=\mathrm{A\,m^{-2}}},
\qquad
\boxed{[m]=\mathrm{A\,m^2}}.
$$
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