indefinite integral of 1/x where x is dimensional?

Question

As we know, the integral of $\frac{1}{x}$ is $ln(x)+c$. Because $x$ and $dx$ have the same dimension, $\int\frac{dx}{x}$ is dimensionless. But my problem is: $x$ is dimensional. I've been trained that the natural log of a dimensional quantity is meaningless, and yet here it is.

Furthermore, it seems like this crops up a lot, such as in separable ODE's. E.g. let's say we have $dx/dt=k x$, where $x$ is dimensional, we obtain $ln(x)=rt+c$. Both sides are dimensionless. But then we take the exponential of both sides and get $x=k e^{rt}$ -- suddenly both sides have dimensions, and $k$ has magically changed from $e^c$, which is dimensionless, into dimensional $k$, but where did the dimension come from? It seems like every textbook I've read is sloppy about this.

Answer 1

$\ln{(x)} + C = \ln{(Cx)}$. The dimensions of $C$ are those which are needed to make $Cx$ dimensionless. :)

Answer 2

Both members of $$\frac{\dot x}x=r$$ are in $s^{-1}$ (speed over distance).

After integration

$$\int\frac{dx}x=\log\left(\frac x{x_0}\right)=r(t-t_0),$$

both members are dimensionless, which allows you to take the exponential:

$$\frac x{x_0}=e^{r(t-t_0)}.$$

Of course,

$$x=x_0e^{r(t-t_0)}$$ are in $m$.