Primitive of Reciprocal of x by Power of Root of a x + b

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Theorem

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


Proof

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

$\displaystyle \int x^m \left({a x + b}\right)^n \ \mathrm d x = \frac {-x^{m+1} \left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) b} + \frac {m + n + 2} {\left({n + 1}\right) b} \int x^m \left({a x + b}\right)^{n + 1} \ \mathrm d x$


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

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

$\blacksquare$


Sources