rope extending under own weight - hooke's law

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I am trying to solve following physics exercise but I cannot find my error.

A rope with a cross section $A$, density $\rho$ and a length $l$ is hanging (staticly). I want to calculate the total length the rope extends under its own weight, under the assumption that the density/total weight and cross section stays the same when stretched, and we know hooke's law:

$$\frac{\Delta l}{l} = \frac{F}{AE}$$

So in physicists manner I tried to use following approach: We use a coordinate $x$, with $x=0$ at the bottom end, $x=l$ at the top end of the rope. We consider a infinitesimal length of rope of length $dx$ at the position $x$. The force pulling on this piece is $F = \rho V g = \rho Axg$, so the elongation of this piece of rope is given by

$$\Delta l = \frac{F}{AE} dx = \frac{\rho A xg}{AE} dx = \frac{\rho xg}{E} dx$$

When summing over all those $dx$ we should get an integral for the total elongation $\Delta L$:

$$\Delta L = \int_0^l \Delta l = \int_0^l \frac{\rho x g}{E} dx = \frac{\rho g}{2E} l^2$$

Is this correct so far? The manipulation of those infinitesimal pieces $dx$ etc. always seems very dubious to me, but thats how our physics professors teach it. Is there any ressource that sheds any light on how to work with those?

EDIT: I think the units are not consistent, lets look first at:

$$\left[\frac{\rho}{E}\right] = \frac{kg/m^3}{Pa} = \frac{s^2}{m^2}$$

This would imply $[\Delta L] = s^2$ which doesn't make any sense at all.