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

$\blacksquare$


Sources