# Integral of Natural Logarithm : ln x

 

The steps to calculate the integral of the natural logarithm function : $\displaystyle \int \ln x \; dx$ are presented.

We first rewrite the given intergral as
$\displaystyle \int \ln x \; dx = \int 1 \cdot \ln x \; dx$

Let $u = x$ and $v = \ln x$ whose first derivatives are given by $u' = 1$ and $v' = \dfrac{1}{x}$

Our integral is of the form
$\displaystyle \int \ln x \; dx = \int u' \cdot v \; dx$

Use the integration by parts to write
$\displaystyle \int \ln x \; dx = u v - \int u \cdot v' \; dx$

Substitute $u, v$ and $v'$ to obtain
$\displaystyle = x \ln x - \int x \dfrac {1}{x} \; dx$

Simplify the term on the right
$\displaystyle = x \ln x - \int dx$

and evaluate the integral
$= x \ln x - x + c$
where $c$ is the constant of integration.
Hence
$\displaystyle \int \ln x \; dx = x \ln x - x + c$