Primitive of Root of a x + b over Power of x/Formulation 1

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Theorem

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


Proof

Let:

\(\displaystyle u\) \(=\) \(\displaystyle \sqrt{a x + b}\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d u}{\mathrm d x}\) \(=\) \(\displaystyle \frac a {2 \sqrt{a x + b} }\) Power Rule for Derivatives etc.
\(\displaystyle v\) \(=\) \(\displaystyle \frac {-1} {\left({m - 1}\right) x^{m - 1} }\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d v}{\mathrm d x}\) \(=\) \(\displaystyle \frac 1 {x^m}\) Power Rule for Derivatives


From Integration by Parts:

$\displaystyle \int u \frac {\mathrm d v} {\mathrm d x} \ \mathrm d x = u v - \int v \ \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$

from which:

\(\displaystyle \int \frac {\sqrt{a x + b} } {x^m} \ \mathrm d x\) \(=\) \(\displaystyle \int \sqrt{a x + b} \frac 1 {x^m} \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt{a x + b} \frac {-1} {\left({m - 1}\right) x^{m - 1} } - \int \frac {-1} {\left({m - 1}\right) x^{m - 1} } \frac a {2 \sqrt{a x + b} } \ \mathrm d x\)
\(\displaystyle \) \(=\) \(\displaystyle -\frac {\sqrt{a x + b} } {\left({m - 1}\right) x^{m-1} } + \frac a {2 \left({m - 1}\right)} \int \frac {\mathrm d x} {x^{m - 1} \sqrt{a x + b} }\) Primitive of Constant Multiple of Function

$\blacksquare$


Sources