Primitive of Logarithm of a x over x

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Theorem

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

for $a \ne 0$ and $x \ne 0$.

Proof 1

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

$\blacksquare$

Proof 2

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

$\blacksquare$

Sources

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