Why does integrating f'(x)/f(x) give different results for equivalent expressions?

[Lucy Yeats]

A fraction obviously stays the same if you multiply top and bottom by the same constant. If you integrate two equal functions, surely you should get the same answer? By applying the rule in the title, you can get different answers. I'm very confused.

Example: 2x/x^2=6x/3x^2
The integral of 2x/x^2 is ln(x^2).
The integral of 6x/3x^2 is ln(3x^2)

How come these aren't equal? Is it to do with the +c?

 

[dalcde]

Remember that $\ln (ab) = \ln a + \ln b$. Hence $\ln(3x^2)=\ln(x^2)+\ln(3)$, so the two expressions just differ by a constant.

 

[Lucy Yeats]

Ah, okay. Thanks!

 

[dimension10]

A fraction obviously stays the same if you multiply top and bottom by the same constant. If you integrate two equal functions, surely you should get the same answer? By applying the rule in the title, you can get different answers. I'm very confused.

Example: 2x/x^2=6x/3x^2
The integral of 2x/x^2 is ln(x^2).
The integral of 6x/3x^2 is ln(3x^2)

How come these aren't equal? Is it to do with the +c?

Yes, it is about the plus C.
ln(3x^2)=ln(3)+ln(x^2)