Why is the integral of 1/x equal to ln absolute x?

I know that:

But when I first learned it, they wrote $\ln x + C$. Then later they added the absolute value. Why?

If $x > 0$, then $\ln|x| = \ln x$ so it doesnt matter. But if $x$ can be negative, $\ln x$ isnt even defined (in the reals).

But consider the definite integral:

Using $\ln|x|$ gives $\ln|-1| - \ln|-2| = \ln 1 - \ln 2 = -\ln 2$, which is correct!

But if we used $\ln x$, wed get $\ln(-1) - \ln(-2)$ which is undefined.

So the question is: why does $\frac{d}{dx} \ln|x| = \frac{1}{x}$? The derivative of $\ln x$ only works for $x > 0$, so how do we handle the negative case?