Primitive of Power of x by Root of a x + b
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Theorem
- $\ds \int x^m \sqrt {a x + b} \rd x = \frac {2 x^m} {\paren {2 m + 3} a} \paren {\sqrt {a x + b} }^3 - \frac {2 m b} {\paren {2 m + 3} a} \int x^{m - 1} \sqrt{a x + b} \rd x$
Proof
From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$:
- $\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$
Putting $n := \dfrac 1 2$:
\(\ds \int x^m \sqrt{a x + b} \rd x\) | \(=\) | \(\ds \int x^m \paren {a x + b}^{1/2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^m \paren {a x + b}^{3/2} } {\paren {m + \frac 1 2 + 1} a} - \frac {m b} {\paren {m + \frac 1 2 + 1} a} \int x^{m - 1} \paren {a x + b}^{1/2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^m \paren {\sqrt {a x + b} }^3} {\paren {m + \frac 3 2} a} - \frac {m b} {\paren {m + \frac 3 2} a} \int x^{m - 1} \sqrt {a x + b} \rd x\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 x^m} {\paren {2 m + 3} a} \paren {\sqrt {a x + b} }^3 - \frac {2 m b} {\paren {2 m + 3} a} \int x^{m - 1} \sqrt {a x + b} \rd x\) | multiplying top and bottom by $2$ |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.96$