Primitive of Power of Root of a x + b/Proof 2

Theorem

$\ds \int \paren {\sqrt {a x + b} }^m \rd x = \frac {2 \paren {\sqrt {a x + b} }^{m + 2} } {a \paren {m + 2} } + C$

for $m \ne -2$.

Let $u = a x + b$.

Then:

$\ds \int \paren {\sqrt {a x + b} }^m \rd x$

$=$

$\ds \frac 1 a \int \paren {\sqrt u}^m \rd u$

$\ds $

$=$

$\ds \frac 1 a \int u^{m / 2} \rd u$

simplifying

$\ds $

$=$

$\ds \frac 1 a \frac {u^{m / 2 + 1} } {m / 2 + 1} + C$

Primitive of Power valid for $m / 2 + 1 \ne 0$, that is, $m \ne -2$

$\ds $

$=$

$\ds \frac 1 a \frac {u^{\paren {m + 2} / 2} } {\paren {m + 2} / 2} + C$

$\ds $

$=$

$\ds \frac 2 a \frac {\paren {\sqrt u}^{m + 2} } {\paren {m + 2} } + C$

$\ds $

$=$

$\ds \frac {2 \paren {\sqrt {a x + b} }^{m + 2} } {a \paren {m + 2} } + C$

substituting for $u$

$\blacksquare$