If one considers the Schwarzschild metric
$$ \text d s^2 = -V(r)\text d t^2 + \frac{1}{V(r)}\text d r^2 + r^2 \text d \Omega^2\;,\qquad V(r) = 1-\frac{2m}{r}\;, $$
and introduces the Eddington-Finkelstein coordinates
$$ v = t+f(r)\;,\qquad u = t-f(r)\;, \qquad f'(r) = \frac{1}{V(r)} \;. $$
the coordinate singularity at $r=2m$ vanishes and the metric reads:
$$ \text d s^2 = -V(r)\text d v\text d u + r^2\text d \Omega^2\,. $$
However, by integrating $f'$ to get $f$ I would calculate $$ f(r) = \int \frac{1}{V(r)}\text d r = \int \left( 1+\frac{2m}{r-2m} \right) d r = r + 2m\ln|r-2m| \;. $$
But in books and lecture notes I always read $\ln(r-2m)$ instead of $\ln|r-2m|$. Isn't that wrong? In my opinion this makes a difference.
For example when I introduce the Kruskal-Szekeres coordinates $$ \tan V = \text e ^{\alpha v}\;,\qquad \tan U = -\text e ^{-\alpha u}\;,\qquad \alpha = \frac{1}{4m}\, $$ I get $$ \tan V \tan U = -\text e ^{\frac{r}{2m}} |r-2m| $$ instead of $$ \tan V \tan U = -\text e ^{\frac{r}{2m}} (r-2m) $$ But these two cases lead to different Penrose diagrams.
So why is it wrong to to use the absolute value of $r-2m$ in the logarithm?
PS: When it is about the Reissner-Nordstrom metric all books and notes I find surprisingly use the magnitude in the argument of the logarithm. But why not in the case of Schwarzschild?