Primitive of Root of a x + b over Power of x/Formulation 2

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Theorem

$\ds \int \frac {\sqrt{a x + b} } {x^m} \rd x = -\frac {\paren {\sqrt{a x + b} }^3} {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 5} a} {\paren {2 m - 2} b} \int \frac {\sqrt {a x + b} } {x^{m - 1} } \rd x$


Proof

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

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


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

\(\ds \int \frac {\sqrt {a x + b} } {x^m} \rd x\) \(=\) \(\ds \int x^{-m} \paren {a x + b}^{1 / 2} \rd x\)
\(\ds \) \(=\) \(\ds \frac {x^{-m + 1} \paren {a x + b}^{1 / 2 + 1} } {\paren {-m + 1} b} - \frac {\paren {-m + \frac 1 2 + 2} a} {\paren {-m + 1} b} \int x^{-m + 1} \paren {a x + b}^{1 / 2} \rd x\)
\(\ds \) \(=\) \(\ds -\frac {\paren {\sqrt {a x + b} }^3} {\paren {m - 1} b x^{m - 1} } - \frac {\paren {m - \frac 5 2} a} {\paren {m - 1} b} \int \frac {\sqrt {a x + b} } {x^{m - 1} } \rd x\) simplifying
\(\ds \) \(=\) \(\ds -\frac {\paren {\sqrt {a x + b} }^3} {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 5} a} {\paren {2 m - 2} b} \int \frac {\sqrt {a x + b} } {x^{m - 1} } \rd x\) multiplying top and bottom by $2$

$\blacksquare$


Sources

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