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:

\(\ds \int \dfrac {\d u} {\d x} \dfrac 1 {\sqrt {\map f x} } \rd x\)

\(=\)

\(\ds \int \dfrac {\d u} {\sqrt u}\)

Integration by Substitution

\(\ds \)

\(=\)

\(\ds 2 \sqrt u + C\)

Primitive of Power

\(\ds \)

\(=\)

\(\ds 2 \sqrt {\map f x} + C\)

Primitive of Power

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Integration