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

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Theorem

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


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 $n = -\dfrac 1 2$:

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

$\blacksquare$


Sources

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