= Solution
First retain the question's units $c=1$. Insert the <retarded time> <Dirac delta>:
$$
\phi(t,x)=\frac1{4\pi\epsilon_0}\int d^3x'\,dt'\,
\frac{\rho(t',x')}{|x-x'|}\delta(t-t'-|x-x'|).
$$
For a <point charge>, the spatial delta sets $x'=x_0(t')$. Put $R(t')=|x-x_0(t')|$, $\widehat R=(x-x_0)/R$ and $v=\dot x_0(t')$. The retarded root $t_*$ satisfies $t-t_*=R(t_*)$. The derivative of the delta's argument is $-1+\widehat R\cdot v$, so for a subluminal trajectory its absolute value is $1-\widehat R\cdot v$. Performing the time integral gives
$$
\boxed{\phi(t,x)=\frac q{4\pi\epsilon_0}\frac1{R_*-v_*\cdot\mathbf R_*}},\qquad
\mathbf R_*=x-x_0(t_*),\quad R_*=|\mathbf R_*|.
$$
All quantities in the denominator are evaluated at the <retarded time>. The retarded <vector potential> uses current $j=qv\delta^{(3)}(x'-x_0)$ and the same <Jacobian determinant>, giving
$$
\boxed{\mathbf A(t,x)=\frac{\mu_0q}{4\pi}\frac{v_*}{R_*-v_*\cdot\mathbf R_*}}\quad(c=1).
$$
Restoring physical $c$, the root is $t-t_*=R_*/c$, both denominators become $R_*-v_*\cdot\mathbf R_*/c$, and $\mu_0=1/(\epsilon_0c^2)$. Thus $\mathbf A=v_*\phi/c^2$. This is the <Liénard–Wiechert potential>; the <Jacobian determinant>, rather than merely substituting retarded position into a static Coulomb formula, accounts for the velocity factor.
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