This is the Biot-Savart Law for the vector potential for magnetostatics. The derivation starts from the static Maxwell equations for the magnetic field (here written in Heaviside-Lorentz units)
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{B}=\frac{\mu}{c} \vec{j}.$$
From the first equation we can write
$$\vec{B}=\vec{\nabla} \times \vec{A},$$
but ##\vec{A}## is defined only up to a gradient field (gauge invariance for the special case of magnetostatics). This can be used to impose one additional condition on ##\vec{A}##. As we shall see in a moment, the following Coulomb-gauge condition is particularly convenient in this case:
$$\vec{\nabla} \cdot \vec{A}=0.$$
Now we use the inhomogeneous equation (Ampere's Law):
$$\vec{\nabla} \times \vec{B}=\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}=\frac{\mu}{c} \vec{j}.$$
Now using the Coulomb-gauge condition this simplifies finally to
$$\Delta \vec{A}=-\frac{\mu}{c} \vec{j}.$$
This is the same equation as for the electrostatic potential, just for every Cartesian component of ##\vec{A}##. Thus we know the solution via the superposition of Coulomb fields (more formally it's the use of the Green's function of the Laplace operator):
$$\vec{A}(\vec{x})=\frac{\mu}{4 \pi c} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\vec{j}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
Note that this is only consistent, if
$$\vec{\nabla} \cdot \vec{j}=0,$$
which is charge conservation for the static case. If this integrability condition (which follows also from the Maxwell equation (Ampere's Law)) is fulfilled, then the found solution also fulfills the Coulomb-gauge condition, as it must be for consistency.