# Why the modulus signs when integrating f'(x)/f(x)?

1. Jan 3, 2005

### Cheman

I am able to proove to myself, through generalised substitution, that the integral of f'(x)/f(x) is lnf(x)+c, but where do the modulus signs come from? ie - The accepted integral is ln|f(x)|+c, not lnf(x)+c

2. Jan 3, 2005

### brendan_foo

I cannot be specific in my answer, but I know that a negative argument will not work with the natural logarithm function, therefore I guess the magnitude of the answer is the argument that is valid.

3. Jan 3, 2005

### HallsofIvy

Staff Emeritus
logarithms are only defined for positive arguments: loga(x) is the inverse to ax and (for a positive) ax is always positive.

But d(ln(x))/dx= 1/x for x positive, and using the chain rule, d(ln(-x))/dx= (1/(-x))(-1)= 1/x with x negative. Thus: the anti-derivative for ln(x) is properly ln|x|+ C rather than ln(x)+ C.

I will confess that I always forget the "| |" myself. Most of the time it doesn't matter: $\int_a^b (1/x)dx= ln b- ln a$ if a and b are both positive,
ln(|b|)- ln(|a|)= ln(-b)- ln(-a) if a and b are both negative so you can just 'ignore' the negative signs. Of course, 1/x is not defined for x= 0 and the integral is not defined if a is negative and b positive.

Last edited: Jan 3, 2005