Natural Log (ln(x)) - When integrating a function, and the answer is ln(x), do we take the absolute value of x?

example
I have a Definite* Integral (3, 4) of [(1/(x-8))dx] - Definite Integral (3,4) of [(1/(x+9)]dx

The answer, I know, comes out to be ln(x-8) - ln(x+9).

so using the law of log, I can make that ln((x-8)/(x+9)

What do I do with the negative values? is the antiderivative supposed to be the absolute value of ln(x) *i.e. ln|x|*?

Question

Natural Log (ln(x)) - When integrating a function, and the answer is ln(x), do we take the absolute value of x? 

example 
I have a Definite* Integral (3, 4) of [(1/(x-8))dx] - Definite Integral (3,4) of [(1/(x+9)]dx 

The answer, I know, comes out to be ln(x-8) - ln(x+9).

so using the law of log, I can make that ln((x-8)/(x+9)

What do I do with the negative values? is the antiderivative supposed to be the absolute value of ln(x) *i.e. ln|x|*?

anonymous · Answer

so i have ln(-48/-65)?

dumbcow · Answer

yes put the abs value inside log to take care of neg values

dumbcow · Answer

so yes answer is ln(48/65)

anonymous · Answer

does ln(48/65) = -ln(65/46)? @dumbcow

dumbcow · Answer

yes because of log property
$$\ln(x^{-1}) = -\ln(x)$$

anonymous · Answer

okay, great.. i fully understand it now. thanks!