in the context of Magnetic dipole expansion I'm analyzing this equation
$$\vec{A} (\vec{x}) = \frac{\mu_0}{4\pi} \frac{1}{\|\vec{x}\|} \left(\int d^3x' \, \vec{j}(\vec{x}') + \frac{1}{\|\vec{x}\|^3} \int d^3x'\,(\vec{x} \cdot \vec{x}')\,\vec{j}(\vec{x}')+...\right)$$
in where the first term (magnetic monopole) is 0, at least in an electrostatic case in which currents are static ($\mathrm{div} \,\vec{j} =0$) and the boundary term also cancels: $\vec{j}\,|_{\partial V} =0$.
For the second term my lecturer states that:
$$\int d^3x' (\vec{x}\cdot\vec{x}')\,\vec{j}(\vec{x}') = \vec{x} \cdot \int d^3 \,\vec{x}' j_i$$
I'm almost sure that this is consequence of commutation of the dot product but I'd like to ratify if this is actually what's happening. I post it here on Math SE because my doubt, even if framed inside the electromagnetic context, is about integration and vector calculus.
I'm assuming that $\vec{x}\cdot$ term can go out the integral as it's applying to a $d^3x'$ volume, then the product $(\vec{x}'\cdot \vec{x}) \vec{j}$ is equivalent to $\vec{x} \cdot (j_i \vec{x}')$ where i is denoting each component of $\vec{j}$.
Does this make sense to you? Is there any limitation, assuming the fields do not diverge nor do anything utterly strange (a physical case where everything is happy)?
Let $u\in \mathbb{R^3}$ be constant and $v:\mathbb{R^3}\rightarrow\mathbb{R^3} .$ Observe
$$ \int\int\int_\Omega u \cdot v(x',y',z') dx' dy'dz' = \int\int\int_\Omega u_1 v_1(x',y',z') + u_2 v_2(x',y',z') + u_3 v_3(x',y',z') dx' dy'dz' $$
thus we have that
$$\begin{align} \int\int\int_\Omega u \cdot v(x',y',z') dx' dy'dz' &= \\ \\u_1\int\int\int_\Omega v_1(x',y',z') dx' dy'dz' &+ u_2\int\int\int_\Omega v_2(x',y',z') dx' dy'dz' \\&+ u_3\int\int\int_\Omega v_3(x',y',z') dx' dy'dz' \end{align}$$
because $u_1,u_2,u_3$ are constant. It is then clear that
$$\begin{align}\int\int\int_\Omega u \cdot v(x',y',z') dx' dy'dz' &= \\ \\ u \cdot \bigg(\int\int\int_\Omega v_1(x',y',z') dx' dy'dz' &+ \int\int\int_\Omega v_2(x',y',z') dx' dy'dz' \\&+ \int\int\int_\Omega v_3(x',y',z') dx' dy'dz'\bigg) \end{align}$$
so we finally achieve that
$$\int\int\int_\Omega u \cdot v(x',y',z') dx' dy'dz' = u \cdot \int\int\int_\Omega v(x',y',z') dx' dy'dz' .$$
Applying this to your problem, if the primed and unprimed coordinates are independent of one another then $\vec{x}$ can be pulled out of the integral in the same fashion.
Let me know if this needs more explaining.