Primitive of Reciprocal of Power of p x + q by Root of a x + b

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Theorem

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


Proof

From Primitive of Reciprocal of $\dfrac 1 {\left({a x + b}\right)^m \left({p x + q}\right)^n}$:

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


Setting $m := \dfrac 1 2$:

\(\displaystyle \int \frac {\mathrm d x} {\left({p x + q}\right)^n \sqrt{a x + b} }\) \(=\) \(\displaystyle \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac 1 {\left({a x + b}\right)^{1/2 - 1} \left({p x + q}\right)^{n-1} } + a \left({\frac 1 2 + n - 2}\right) \int \frac {\mathrm d x} {\left({a x + b}\right)^{1/2} \left({p x + q}\right)^{n-1} } }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sqrt{a x + b} } {\left({n - 1}\right) \left({a q - b p}\right) \left({p x + q}\right)^{n-1} } + \frac {\left({2 n - 3}\right) a} {2 \left({n - 1}\right) \left({a q - b p}\right)} \int \frac {\mathrm d x} {\left({p x + q}\right)^{n-1} \sqrt{a x + b} }\)

$\blacksquare$


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