Primitive of Reciprocal of x squared by Root of a x + b

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Theorem

$\ds \int \frac {\d x} {x^2 \sqrt {a x + b} } = -\frac {\sqrt {a x + b} } {b x} - \frac a {2 b} \int \frac {\d x} {x \sqrt {a x + b} }$


Proof

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

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

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

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

$\blacksquare$


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