Primitive of x by Power of Root of a x + b
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Theorem
- $\ds \int x \paren {\sqrt {a x + b} }^m \rd x = \frac {2 \paren {\sqrt {a x + b} }^{m + 4} } {a^2 \paren {m + 4} } - \frac {2 b \paren {\sqrt {a x + b} }^{m + 2} } {a^2 \paren {m + 2} } + C$
Proof
Let:
\(\ds u\) | \(=\) | \(\ds \sqrt {a x + b}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \frac {u^2 - b} a\) |
Then:
\(\ds \int x \paren {\sqrt {a x + b} }^m \rd x\) | \(=\) | \(\ds \frac 2 a \int \frac {u^2 - b} a u^{m + 1} \rd x\) | Primitive of Function of $a x + b$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {a^2} \int u^{m + 3} \rd x - \frac {2 b} {a^2} \int u^{m + 1} \rd x\) | Linear Combination of Primitives | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {a^2} \frac {u^{m + 4} } {m + 4} - \frac {2 b} {a^2} \frac {u^{m + 2} } {m + 2} + C\) | Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \paren {\sqrt {a x + b} }^{m + 4} } {a^2 \paren {m + 4} } - \frac {2 b \paren {\sqrt {a x + b} }^{m + 2} } {a^2 \paren {m + 2} } + C\) | substituting for $u$ |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.100$