Primitive of Root of a x + b over Power of x/Formulation 2

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Theorem

$\displaystyle \int \frac {\sqrt{a x + b} } {x^m} \ \mathrm d x = -\frac {\left({\sqrt{a x + b} }\right)^3} {\left({m - 1}\right) b x^{m-1} } - \frac {\left({2 m - 5}\right) a} {\left({2 m - 2}\right) b} \int \frac {\sqrt{a x + b} } {x^{m - 1} } \ \mathrm d x$


Proof

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

$\displaystyle \int x^m \left({a x + b}\right)^n \ \mathrm d x = \frac {x^{m+1} \left({a x + b}\right)^{n + 1} } {\left({m + 1}\right) b} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \ \mathrm d x$


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

\(\displaystyle \int \frac {\sqrt{a x + b} } {x^m} \ \mathrm d x\) \(=\) \(\displaystyle \int x^{-m} \left({a x + b}\right)^{1/2} \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^{-m+1} \left({a x + b}\right)^{1/2 + 1} } {\left({-m + 1}\right) b} - \frac {\left({-m + \frac 1 2 + 2}\right) a} {\left({-m + 1}\right) b} \int x^{-m + 1} \left({a x + b}\right)^{1/2} \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac {\left({\sqrt{a x + b} }\right)^3} {\left({m - 1}\right) b x^{m-1} } - \frac {\left({m - \frac 5 2}\right) a} {\left({m - 1}\right) b} \int \frac {\sqrt{a x + b} } {x^{m - 1} } \ \mathrm d x\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle -\frac {\left({\sqrt{a x + b} }\right)^3} {\left({m - 1}\right) b x^{m-1} } - \frac {\left({2 m - 5}\right) a} {\left({2 m - 2}\right) b} \int \frac {\sqrt{a x + b} } {x^{m - 1} } \ \mathrm d x\) multiplying top and bottom by $2$

$\blacksquare$


Sources