Primitive of Function under its Derivative

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Then:

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

where $C$ is an arbitrary constant.


Proof

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

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

Then:

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

$\blacksquare$


Also presented as

This can also be presented as:

$\ds \int \frac {\d u} u = \ln \size u + C$

where it is understood that $u$ is a function of $x$.


Sources