Primitive of x by Root of a x + b
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Theorem
- $\ds \int x \sqrt {a x + b} \rd x = \frac {2 \paren {3 a x - 2 b} } {15 a^2} \sqrt {\paren {a x + b}^3}$
Proof
Let:
\(\ds u\) | \(=\) | \(\ds \sqrt {a x + b}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \frac {u^2 - b} a\) |
Then:
\(\ds \int x \sqrt {a x + b} \rd x\) | \(=\) | \(\ds \frac 2 a \int \frac {u^2 - b} a u^2 \rd u\) | Primitive of Function of $\sqrt {a x + b}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {a^2} \int u^4 \rd u - \frac {2 b} {a^2} \int u^2 \rd u\) | Linear Combination of Primitives | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {a^2} \frac {u^5} 5 - \frac {2 b} {a^2} \frac {u^3} 3 + C\) | Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {6 u^5 - 10 b u^3} {15 a^2} + C\) | common denominator | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {6 u^2 - 10 b} u^3} {15 a^2} + C\) | extracting $u^3$ as a factor | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {6 \paren {a x + b} - 10 b} {15 a^2} \sqrt {\paren {a x + b}^3} + C\) | substituting for $u$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \paren {3 a x - 2 b} } {15 a^2} \sqrt {\paren {a x + b}^3} + C\) | simplifying |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.90$