Primitive of Reciprocal

Theorem

$\ds \int \frac {\d x} x = \ln \size x + C$

for $x \ne 0$.

Corollary 1

$\displaystyle \int \frac {\d x} x = \ln x + C$

for $x > 0$.

Corollary 2

$\displaystyle \frac {\d} {\d x} \ln \size x = \frac 1 x$

for $x \ne 0$.

Proof

Suppose $x > 0$.

Then:

$\ln \size x = \ln x$

The result follows from Derivative of Natural Logarithm Function and the definition of primitive.

Suppose $x < 0$.

Then:

 $\ds \dfrac \d {\d x} \ln \size x$ $=$ $\ds \dfrac \d {\d x} \map \ln {-x}$ Definition of Absolute Value $\ds$ $=$ $\ds \frac 1 {-x} \cdot -1$ Chain Rule for Derivatives and Derivative of Natural Logarithm Function, as $-x > 0$ $\ds$ $=$ $\ds \frac 1 x$

and the result again follows from the definition of the primitive.

$\blacksquare$

Also presented as

Some sources gloss over the case where $x < 0$ and merely present this result as:

$\ds \int \frac {\d x} x = \ln x + C$

As far as $\mathsf{Pr} \infty \mathsf{fWiki}$ is concerned, that is a mistake.

Sources

but beware that it reports $\ds \int \frac {\d x} x = \ln x$