Primitive of Root of Function under Half its Derivative

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

Then:


 * $\ds \int \frac {\map {f'} x} {2 \sqrt {\map f x} } \rd x = \sqrt {\map f x} + C$

where $C$ is an arbitrary constant.

Proof
By Integration by Substitution (with appropriate renaming of variables):
 * $\ds \int \map g u \rd u = \int \map g {\map f x} \map {f'} x \rd x$

Let $\map u x = \sqrt {\map f x}$

Then: