Square Root of Function under Derivative

Theorem

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

Proof
Let $u = \map f x$.

Then:
 * $\dfrac {\d u} {\d x} = \map {f'} x$

Hence: