Primitive of Root of Function under Half its Derivative

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

\(\ds \map u x\) \(=\) \(\ds \sqrt {\map f x}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\d u} {\d x}\) \(=\) \(\ds \dfrac {\map {f'} x} {2 \sqrt {\map f x} }\) Chain Rule for Derivatives, Derivative of Power


Then:

\(\ds \int \frac {\map {f'} x} {2 \sqrt {\map f x} } \rd x\) \(=\) \(\ds \int \rd u\) Integration by Substitution: putting $\map u x = \sqrt {\map f x}$
\(\ds \) \(=\) \(\ds u + C\) Primitive of Constant
\(\ds \) \(=\) \(\ds \sqrt {\map f x} + C\) Definition of $u$

$\blacksquare$


Sources