Primitive of Logarithm of x over x

Theorem

$\ds \int \frac {\ln x} x \rd x = \frac {\ln^2 x} 2 + C$

$\ds \int \frac {\ln a x} x \rd x = \frac {\paren {\ln a x}^2} 2 + C$

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 \ \ $

$\ds \frac {\d u} {\d x}$

$=$

$\ds \frac 1 x$

and let:

$\ds \frac {\d v} {\d x}$

$=$

$\ds \frac 1 x$

$\ds \leadsto \ \ $

$\ds v$

$=$

$\ds \ln x$

Then:

$\ds \int \frac {\ln x} x \rd x$

$=$

$\ds \ln x \ln x - \int \ln x \paren {\frac 1 x} \rd x + C$

$\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

$\blacksquare$

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