Primitive of Function under its Derivative

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 $x$.