= Solution
Use the identity $\nabla\times\mathbf B=\nabla(\nabla\cdot\mathbf A)-\nabla^2\mathbf A$. The <Laplacian> identity for the Coulomb <kernel> gives, component by component,
$$
-\nabla^2\mathbf A=-\frac{\mu_0}{4\pi}\int\mathbf J(\mathbf r')\nabla^2(s^{-1})\,dV'=\mu_0\mathbf J(\mathbf r).
$$
Also $\nabla(s^{-1})=-\nabla'(s^{-1})$, so <integration by parts> yields
$$
\begin{aligned}\nabla\cdot\mathbf A&=-\frac{\mu_0}{4\pi}\int\mathbf J(\mathbf r')\cdot\nabla'(s^{-1})\,dV'\\&=\frac{\mu_0}{4\pi}\int\frac{\nabla'\cdot\mathbf J(\mathbf r')}{s}\,dV'.\end{aligned}
$$
Here the surface integral at infinity vanishes under the assumed localization of the current. As usual, this assumption means sufficient decay for these integrals and the boundary limit, rather than merely pointwise convergence to zero. Combining the two terms gives
$$
\boxed{\nabla\times\mathbf B=\mu_0\mathbf J+\frac{\mu_0}{4\pi}\nabla\int\frac{\nabla'\cdot\mathbf J(\mathbf r')}{|\mathbf r-\mathbf r'|}\,dV'.}
$$
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