Primitive of Power of Root of a x + b

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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 1

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 $u$

$\blacksquare$


Proof 2

Let $u = a x + b$.

Then:

\(\ds \int \paren {\sqrt {a x + b} }^m \rd x\) \(=\) \(\ds \frac 1 a \int \paren {\sqrt u}^m \rd u\) Primitive of Function of $a x + b$
\(\ds \) \(=\) \(\ds \frac 1 a \int u^{m / 2} \rd u\) 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 $u$

$\blacksquare$


Also presented as

This result is also seen presented in the form:

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


Sources