Assume $c = 1$ for what follows.
For the general inhomogenous wave equation in one spatial dimension $$\left(\frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2}\right)\phi = v(x, t),$$ the article The Wave Equation with a Source gives (where we ignore the homogenous part of the solution, e.g. by setting initial conditions to $0$): $$\phi(x, t) = \frac{1}{2} \int_0^t \int_{x - (t - s)}^{x + (t - s)} v(y, s)\ dy \ ds.$$
For the general inhomogenous wave equation in three spatial dimensions $$\left(\frac{\partial^2}{\partial t^2} - \nabla^2\right)\phi = v({\bf r}, t),$$ the article Solution of Inhomogeneous Wave Equation gives $$\phi({\bf r}, t) = \int \frac{ v({\bf r}', t - \vert{\bf r} -{\bf r}'\vert)} {4\pi \vert{\bf r} - {\bf r}'\vert}\ dV'.$$
Why does the integrand in the three-dimensional case depend inversely on $\vert{\bf r} - {\bf r}'\vert$, while the integrand in the one-dimensional case doesn't?