Primitive of Logarithm of a x over x
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Theorem
- $\ds \int \frac {\ln a x} x \rd x = \frac {\paren {\ln a x}^2} 2 + C$
Proof
Let $z = a x$.
| \(\ds z\) | \(=\) | \(\ds a x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \d z\) | \(=\) | \(\ds a \rd x\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \int \frac {\ln a x} x \rd x\) | \(=\) | \(\ds \int \frac {\ln z} {z / a} \dfrac {\rd z} a\) | Integration by Substitution: $z = a x$ | ||||||||||
| \(\ds \) | \(=\) | \(\ds \int \frac {\ln z \rd z} z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\paren {\ln z}^2} 2 + C\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\paren {\ln a x}^2} 2 + C\) |
$\blacksquare$
Sources
- 1968: George B. Thomas, Jr.: Calculus and Analytic Geometry (4th ed.) ... (previous) ... (next): Back endpapers: A Brief Table of Integrals: $111$.