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

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Theorem

$\ds \int \frac {\sqrt {a x + b} } {\paren {p x + q}^n} \rd x = \frac {-\sqrt {a x + b} } {\paren {n - 1} p \paren {p x + q}^{n - 1} } + \frac a {2 \paren {n - 1} p} \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }$


Proof

From Primitive of Power of $a x + b$ over Power of $p x + q$: Formulation 3:

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


Setting $m := \dfrac 1 2$:

\(\ds \int \frac {\sqrt {a x + b} } {\paren {p x + q}^n} \rd x\) \(=\) \(\ds \frac {-1} {\paren {n - 1} p} \paren {\frac {\paren {a x + b}^{1/2} } {\paren {p x + q}^{n - 1} } + \frac 1 2 a \int \frac {\paren {a x + b}^{1/2 - 1} } {\paren {p x + q}^{n - 1} } \rd x}\)
\(\ds \) \(=\) \(\ds \frac {-\sqrt {a x + b} } {\paren {n - 1} p \paren {p x + q}^{n - 1} } + \frac a {2 \paren {n - 1} p} \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }\)

$\blacksquare$


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