# What is the antiderivative of ln x?

Jun 29, 2016

$\int \ln x \mathrm{dx} = x \ln x - x + C$

#### Explanation:

The integral (antiderivative) of $\ln x$ is an interesting one, because the process to find it is not what you'd expect.

We will be using integration by parts to find $\int \ln x \mathrm{dx}$:
$\int u \mathrm{dv} = u v - \int v \mathrm{du}$
Where $u$ and $v$ are functions of $x$.

Here, we let:
$u = \ln x \to \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x} \to \mathrm{du} = \frac{1}{x} \mathrm{dx}$ and $\mathrm{dv} = \mathrm{dx} \to \int \mathrm{dv} = \int \mathrm{dx} \to v = x$

Making necessary substitutions into the integration by parts formula, we have:
$\int \ln x \mathrm{dx} = \left(\ln x\right) \left(x\right) - \int \left(x\right) \left(\frac{1}{x} \mathrm{dx}\right)$
$\to \left(\ln x\right) \left(x\right) - \int \cancel{x} \left(\frac{1}{\cancel{x}} \mathrm{dx}\right)$
$= x \ln x - \int 1 \mathrm{dx}$
$= x \ln x - x + C \to$ (don't forget the constant of integration!)