# Primitive of Function under its Derivative

## 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 $u$

$\blacksquare$