# What is the integration of 1/x?

Apr 11, 2015

$\int \frac{1}{x} \mathrm{dx} = \ln \left\mid x \right\mid + C$

The reason depends on which definition of $\ln x$ you have used.

I prefer:
Definition: $\ln x = {\int}_{1}^{x} \frac{1}{t} \mathrm{dt}$ for $x > 0$

By the Fundamental Theorem of Calculus, we get: $\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}$ for $x > 0$

From that and the chain rule, we also get $\frac{d}{\mathrm{dx}} \left(\ln \left(- x\right)\right) = \frac{1}{x}$ for $x < 0$

On an interval that excludes $0$, the antiderivative of $\frac{1}{x}$ is
$\ln x$ if the interval consists of positive numbers and it is $\ln \left(- x\right)$ if the interval consists of negative numbers.

$\ln \left\mid x \right\mid$ covers both cases.