# 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

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

## 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

$\blacksquare$