Primitive of Function under its Derivative

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Theorem

Let $f$ be a real function which is integrable.


Then:

$\displaystyle \int \frac {\map {f'} x} {\map f x} \rd x = \ln \size {\map f x} + C$

where $C$ is an arbitrary constant.


Proof

By Integration by Substitution (with appropriate renaming of variables):

$\displaystyle \int \map g u \rd u = \int \map g {\map f x} \, \map {f'} x \rd x$

Then:

\(\displaystyle \int \frac {\map {f'} x} {\map f x} \rd x\) \(=\) \(\displaystyle \int \frac {\d u} u\) putting $\map g u := \dfrac 1 u$
\(\displaystyle \) \(=\) \(\displaystyle \ln \size u + C\) Primitive of Reciprocal
\(\displaystyle \) \(=\) \(\displaystyle \ln \size {\map f x} + C\) Definition of $u$

$\blacksquare$


Sources