Primitive of x over Root of a x + b
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Theorem
- $\ds \int \frac {x \rd x} {\sqrt {a x + b} } = \frac {2 \paren {a x - 2 b} \sqrt {a x + b} } {3 a^2}$
Proof
Let:
\(\ds u\) | \(=\) | \(\ds \sqrt {a x + b}\) | ||||||||||||
\(\ds x\) | \(=\) | \(\ds \frac {u^2 - b} a\) |
Thus:
\(\ds \map F {\sqrt {a x + b} }\) | \(=\) | \(\ds \frac x {\sqrt {a x + b} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map F u\) | \(=\) | \(\ds \paren {\frac {u^2 - b} a} \frac 1 u\) |
Then:
\(\ds \int \frac {x \rd x} {\sqrt {a x + b} }\) | \(=\) | \(\ds \frac 2 a \int u \paren {\frac {u^2 - b} a} \frac 1 u \rd u\) | Primitive of Function of $\sqrt {a x + b}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {a^2} \int \paren {u^2 - b} \rd u\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {a^2} \paren {\frac {u^3} 3 - b u} + C\) | Primitive of Power and Primitive of Constant | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {a^2} \paren {\frac {\paren {a x + b} \sqrt {a x + b} } 3 - b \sqrt {a x + b} } + C\) | substituting for $u$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {3 a^2} \paren {\paren {a x + b} - 3 b} \sqrt {a x + b} + C\) | extracting common factors | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \paren {a x - 2 b} \sqrt {a x + b} } {3 a^2} + C\) | simplifying |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x + b}$: $14.85$