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

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Theorem

$\ds \int \frac {\d x} {x \paren {\sqrt {a x + b} }^m} = \frac 2 {\paren {m - 2} b \paren {\sqrt {a x + b} }^{m - 2} } + \frac 1 b \int \frac {\d x} {x \paren {\sqrt {a x + b} }^{m - 2} }$


Proof

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

$\ds \int x^m \paren {a x + b}^n \rd x = \frac {-x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {n + 1} b} + \frac {m + n + 2} {\paren {n + 1} b} \int x^m \paren {a x + b}^{n + 1} \rd x$


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

\(\ds \int \frac {\d x} {x \paren {\sqrt {a x + b} }^m}\) \(=\) \(\ds \int x^{-1} \paren {a x + b}^{-m/2}\)
\(\ds \) \(=\) \(\ds \frac {-x^0 \paren {a x + b}^{-m/2 + 1} } {\paren {-\frac m 2 + 1} b} + \frac {-1 - \frac m 2 + 2} {\paren {-\frac m 2 + 1} b} \int x^{-1} \paren {a x + b}^{- m/2 + 1} \rd x\)
\(\ds \) \(=\) \(\ds \frac 2 {\paren {m - 2} b \paren {a x + b}^{\paren {m - 2} / 2} } - \frac 1 b \int \frac {\d x} {x \paren {a x + b}^{\paren {m - 2} / 2} }\) simplifying
\(\ds \) \(=\) \(\ds \frac 2 {\paren {m - 2} b \paren {\sqrt {a x + b} }^{m - 2} } + \frac 1 b \int \frac {\d x} {x \paren {\sqrt {a x + b} }^{m - 2} }\)

$\blacksquare$


Sources

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