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
\[ \int \ln x \; dx = \int 1 \cdot \ln x \; dx \]
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
\[ \int \ln x \; dx = \int u' \cdot v \; dx \]
Use the integration by parts to write
\[ \int \ln x \; dx = u v - \int u \cdot v' \; dx \]
Substitute \( u, v \) and \( v' \) to obtain
\[ = x \ln x - \int x \dfrac {1}{x} \; dx \]
Simplify the term on the right
\[ = x \ln x - \int dx \]
and evaluate the integral
\[ = x \ln x - x + c \]
where \( c \) is the constant of integration.
Hence
\[ \displaystyle \int \ln x \; dx = x \ln x - x + c \]