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
\[ \displaystyle \int \ln x \; dx = x \ln x - x + c \]