# How to derive this log related integration formula?

gibberingmouther
here is a link with the formula:
https://portal.uea.ac.uk/documents/6207125/8199714/steps+into+calculus+integration+and+natural+logarithms.pdf

i'm talking about the formula that says the integral of f'(x)/f(x)dx = ln(f(x))+C

it's kind of hard to put this into Google. where does this formula come from, and can i derive it using basic math i would know (algebra, calculus, etc.)? any way there is of showing why this formula is true would be appreciated ... i like to see multiple demonstrations if they exist.

Homework Helper
$$\int \frac{f^\prime(x)}{f(x)}\mathrm{d}x=log(x)+C$$
is the same as
$$\int \frac{\mathrm{d}u}{u}=log(u)+C$$
with
u=f(x)
du=f(x)dx
since by the chain rule if g(u)=log(u)
$$g(u)+C=\int g^\prime(u)\mathrm{d}u=\int g^\prime(f(x))f^\prime(x)\mathrm{d}x$$

gibberingmouther
Mentor
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You will have to use something! Easiest would be that ##(e^x)' = e^x## and with the chain rule ##(e^{f(x)})'=e^{f(x)}\cdot f'(x)##.
Now we set ##\log f(x) = \log y = z## which means ##e^z=y##. Our previous equation is then ##(e^z)'=e^z\cdot z'## or likewise ##\dfrac{(e^z)'}{e^z}=z'## which is integrated ##\int \dfrac{(e^z)'}{e^z}\,dx = \int \dfrac{y'}{y}\,dx = \int \dfrac{f(x)'}{f(x)}\,dx = \int z'\,dx = z = \log y = \log f(x)##.

gibberingmouther