Hey people, I just came up across something I didn't know about in my revision. Say I want to find the integral

$\displaystyle \int \frac{1}{3x}dx$

Normally I would do this as following

$\displaystyle \frac{1}{3}\int \frac{1}{x}dx=\frac{1}{3}\ln{x}+k$

However, is it not true that I can also make this of the form

$\displaystyle \int \frac{f'(x)}{f(x)}dx$

via

$\displaystyle \frac{1}{3}\int \frac{3}{3x}dx$

which via a simple u sub yields

$\displaystyle \frac{1}{3}\ln{3x}+j=\frac{1}{3}\ln{x}+\frac{1}{3} \ln{3}+j$

Which does not seem to be equivalent to the original answer. Is it the case that both are valid, but the constants k and j are different? Is k=j+ln3?

Mr F says: They are equivalent. j + ln 3 is just as arbitrary as k .....
Also, how would I go about expressing the answers

$\displaystyle y=\frac{1}{3}\ln{x}+k$ and $\displaystyle \frac{1}{3}\ln{3x}+j$

as explicit functions of x?

Mr F says: They already are explicit functions of x. Do you mean how to express them as explicit functions of y ....?
Ta muchly.