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

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Theorem

$\displaystyle \int \paren {p x + q}^n \sqrt {a x + b} \rd x = \frac {2 \paren {p x + q}^{n + 1} \sqrt {a x + b} } {\paren {2 n + 3} p} + \frac {b p - a q} {\paren {2 n + 3} p} \int \frac {\paren {p x + q}^n} {\sqrt{a x + b} } \rd x$


Proof

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

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


Setting $m := \dfrac 1 2$:

\(\displaystyle \int \paren {a x + b}^{1 / 2} \paren {p x + q}^n \rd x\) \(=\) \(\displaystyle \frac {\paren {a x + b}^{1 / 2} \paren {p x + q}^{n + 1} } {\paren {\frac 1 2 + n + 1} p} + \frac {\dfrac 1 2 \paren {b p - a q} } {\paren {\frac 1 2 + n + 1} p} \int \paren {a x + b}^{1 / 2 - 1} \paren {p x + q}^n \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 \paren {p x + q}^{n + 1} \sqrt {a x + b} } {\paren {2 n + 3} p} + \frac {b p - a q} {\paren {2 n + 3} p} \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x\)

$\blacksquare$


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