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

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Theorem

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


Proof

From Primitive of Power of $a x + b$ over Power of $p x + q$: Formulation 3:

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


Setting $m := \dfrac 1 2$:

\(\displaystyle \int \frac {\sqrt{a x + b} } {\left({p x + q}\right)^n} \ \mathrm d x\) \(=\) \(\displaystyle \frac {-1} {\left({n - 1}\right) p} \left({\frac {\left({a x + b}\right)^{1/2} } {\left({p x + q}\right)^{n-1} } + \frac 1 2 a \int \frac {\left({a x + b}\right)^{1/2-1} } { \left({p x + q}\right)^{n-1} } \ \mathrm d x}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {- \sqrt{a x + b} } {\left({n - 1}\right) p \left({p x + q}\right)^{n-1} } + \frac a {2 \left({n - 1}\right) p} \int \frac {\mathrm d x} {\left({p x + q}\right)^{n-1} \sqrt{a x + b} }\)

$\blacksquare$


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