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

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Theorem

$\ds \int \frac {\sqrt {a x + b} } {x^m} \rd x = -\frac {\sqrt {a x + b} } {\paren {m - 1} x^{m - 1} } + \frac a {2 \paren {m - 1} } \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$


Proof

Let:

\(\ds u\) \(=\) \(\ds \sqrt {a x + b}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds \frac a {2 \sqrt {a x + b} }\) Power Rule for Derivatives etc.
\(\ds v\) \(=\) \(\ds \frac {-1} {\paren {m - 1} x^{m - 1} }\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d v} {\d x}\) \(=\) \(\ds \frac 1 {x^m}\) Power Rule for Derivatives


From Integration by Parts:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

from which:

\(\ds \int \frac {\sqrt {a x + b} } {x^m} \rd x\) \(=\) \(\ds \int \sqrt {a x + b} \frac 1 {x^m} \rd x\)
\(\ds \) \(=\) \(\ds \sqrt {a x + b} \frac {-1} {\paren {m - 1} x^{m - 1} } - \int \frac {-1} {\paren {m - 1} x^{m - 1} } \frac a {2 \sqrt {a x + b} } \rd x\)
\(\ds \) \(=\) \(\ds -\frac {\sqrt {a x + b} } {\paren {m - 1} x^{m - 1} } + \frac a {2 \paren {m - 1} } \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }\) Primitive of Constant Multiple of Function

$\blacksquare$


Sources