Primitive of Power of Root of a x + b over x

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Theorem

$\displaystyle \int \frac {\left({\sqrt{a x + b} }\right)^m} x \ \mathrm d x = \frac {2 \left({\sqrt{a x + b} }\right)^m } m + b \int \frac {\left({\sqrt{a x + b} }\right)^{m - 2} } x \ \mathrm d x$


Proof

From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $a x + b$:

$\displaystyle \int x^m \left({a x + b}\right)^n \ \mathrm d x = \frac {x^{m+1} \left({a x + b}\right)^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \left({a x + b}\right)^{n - 1} \ \mathrm d x$


Putting $n := \dfrac m 2$ and $m := -1$:

\(\displaystyle \int \frac {\left({\sqrt{a x + b} }\right)^m} x \ \mathrm d x\) \(=\) \(\displaystyle \int x^{-1} \left({a x + b}\right)^{m/2} \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^0 \left({a x + b}\right)^{m/2} } {-1 + \frac m 2 + 1} + \frac {\left({\frac m 2}\right) b} {-1 + \frac m 2 + 1} \int x^{-1} \left({a x + b}\right)^{m/2 - 1} \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 \left({a x + b}\right)^{m/2} } m + b \int x^{-1} \left({a x + b}\right)^{\left({m - 2}\right)/2} \ \mathrm d x\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 \left({\sqrt{a x + b} }\right)^m } m + b \int \frac {\left({\sqrt{a x + b} }\right)^{m - 2} } x \ \mathrm d x\)

$\blacksquare$


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