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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.97$