# Integral question

Why is the anti derivative of 1/(x-3) equal to -ln$$\left|x+3\right|$$. Why isn't it just ln$$\left|x-3\right|$$. Does it have something to do with the absolute value? Thanks.

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for negative x?

using just log |x-3| defines the differentiation function for pos x only

exk
it's not -ln|x+3| and is in fact ln|x-3| . Take the derivative of your first answer and second and see which one works.

Alright so I'm guessing there was a typo. Thanks for your help.

d/dx (log |x-3|) != 1/(x-3)
for x element of (-inf, inf) for sure! edit: wrong

ooops.. my mistake. I was ignoring the absolute part

Last edited:
rock.freak667
Homework Helper
d/dx (log |x-3|) != 1/(x-3)
for x element of (-inf, inf) for sure!
For $x \neq 3$

exk
the original function is undefined for x=3 also btw.