Primitive of Root of a x + b over x squared

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Theorem

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


Proof

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

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


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

\(\ds \int \frac {\sqrt {a x + b} } {x^2} \rd x\) \(=\) \(\ds \int x^{-2} \paren {a x + b}^{1/2} \rd x\)
\(\ds \) \(=\) \(\ds \frac {x^{-1} \paren {a x + b}^{1/2} } {-2 + \frac 1 2 + 1} + \frac {\frac 1 2 b} {-2 + \frac 1 2 + 1} \int x^{-2} \paren {a x + b}^{-1/2} \rd x\)
\(\ds \) \(=\) \(\ds \frac {-2 \sqrt {a x + b} } x - b \int \frac {\d x} {x^2 \sqrt {a x + b} }\) simplifying
\(\ds \) \(=\) \(\ds \frac {-2 \sqrt {a x + b} } x - b \paren {-\frac {\sqrt {a x + b} } {b x} - \frac a {2 b} \int \frac {\d x} {x \sqrt {a x + b} } }\) Primitive of Reciprocal of x squared by Root of a x + b
\(\ds \) \(=\) \(\ds \frac {-2 \sqrt {a x + b} } x + \frac {\sqrt {a x + b} } x + \frac a 2 \int \frac {\d x} {x \sqrt {a x + b} }\) multiplying out
\(\ds \) \(=\) \(\ds -\frac {\sqrt {a x + b} } x + \frac a 2 \int \frac {\d x} {x \sqrt {a x + b} }\) simplifying

$\blacksquare$


Also see


Sources

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