Primitive of Logarithm of x over x
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Theorem
- $\ds \int \frac {\ln x} x \rd x = \frac {\ln^2 x} 2 + C$
Corollary
- $\ds \int \frac {\ln a x} x \rd x = \frac {\paren {\ln a x}^2} 2 + C$
Proof
With a view to expressing the primitive in the form:
- $\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
\(\ds u\) | \(=\) | \(\ds \ln x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds \frac 1 x\) | Derivative of $\ln x$ |
and let:
\(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds \frac 1 x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds \ln x\) | Primitive of Reciprocal |
Then:
\(\ds \int \frac {\ln x} x \rd x\) | \(=\) | \(\ds \ln x \ln x - \int \ln x \paren {\frac 1 x} \rd x + C\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \ln^2 x - \int \frac {\ln x} x \rd x + C\) | tidying | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 \int \frac {\ln x} x \rd x\) | \(=\) | \(\ds \ln^2 x + C\) | adding $\ds\int \frac {\ln x} x \rd x$ to both sides | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {\ln x} x \rd x\) | \(=\) | \(\ds \frac {\ln^2 x} 2 + C\) | simplifying |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\ln x$: $14.528$
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Appendix: Table $2$: Integrals
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Appendix: Table $2$: Integrals