Primitive of Function of Root of a x + b

Theorem

$\ds \int \map F {\sqrt {a x + b} } \rd x = \frac 2 a \int u \map F u \rd u$

where $u = \sqrt {a x + b}$.

$\ds u$

$=$

$\ds \sqrt {a x + b}$

$\ds u$

$=$

$\ds \paren {a x + b}^{1/2}$

$\ds \leadsto \ \ $

$\ds \frac {\d u} {\d x}$

$=$

$\ds \frac 1 2 \paren {a x + b}^{-1/2} \map {\frac \d {\d x} } {a x + b}$

$\ds $

$=$

$\ds \frac 1 {2 u} \map {\frac \d {\d x} } {a x + b}$

substituting for $u$

$\ds $

$=$

$\ds \frac a {2 u}$

$\ds \leadsto \ \ $

$\ds \int \map F {\sqrt {a x + b} } \rd x$

$=$

$\ds \int \frac {2 u} a \map F u \rd u$

$\ds $

$=$

$\ds \frac 2 a \int u \map F u \rd u$

$\blacksquare$