Primitive of Reciprocal of Root of a x + b
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Theorem
- $\ds \int \frac {\d x} {\sqrt{a x + b} } = \frac {2 \sqrt {a x + b} } a + C$
where $a x + b > 0$.
Proof
Put $u = \sqrt{a x + b}$.
Then:
\(\ds \int \frac {\d x} {\sqrt {a x + b} }\) | \(=\) | \(\ds \frac 2 a \int u \frac {\d u} u\) | Primitive of Function of $\sqrt {a x + b}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 a \int \rd u\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 a u + C\) | Primitive of Constant | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \sqrt {a x + b} } a + C\) | substituting for $u$ |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt{a x + b}$: $14.84$
- 1976: K. Weltner and W.J. Weber: Mathematics for Engineers and Scientists ... (previous) ... (next): $6$. Integral Calculus: Appendix: Table of Fundamental Standard Integrals