Primitive of Power of Root of a x + b/Proof 1

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 $m \ne -2$.

Proof

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

Then:

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

\(=\)

\(\ds \frac 2 a \int u \cdot u^m \rd x\)

Primitive of Function of $\sqrt {a x + b}$

\(\ds \)

\(=\)

\(\ds \frac 2 a \int u^{m + 1} \rd x\)

simplifying

\(\ds \)

\(=\)

\(\ds \frac 2 a \frac {u^{m + 2} } {m + 2} + C\)

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

\(\ds \)

\(=\)

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

substituting for $u$

$\blacksquare$