Integration of One over X

The integration of one over $$x$$ is another important formula of integration.

The integration of one over $$X$$ is of the form
\[\int {\frac{1}{x}dx = } \ln x + c\]

Now consider
\[\begin{gathered} \frac{d}{{dx}}\left[ {\ln x + c} \right] = \frac{d}{{dx}}\ln x + \frac{d}{{dx}}\left( c \right) \\ \Rightarrow \frac{d}{{dx}}\left[ {\ln x + c} \right] = \frac{1}{x} + 0 \\ \Rightarrow \frac{1}{x} = \frac{d}{{dx}}\left[ {\ln x + c} \right] \\ \Rightarrow \frac{1}{x}dx = d\left[ {\ln x + c} \right]\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ \end{gathered} \]

Integrating both sides of equation (i) with respect to $$x$$, we have
\[\int {\frac{1}{x}dx} = \int {d\left[ {\ln x + c} \right]} \]

Since integration and differentiation are reverse processes to each other, the integral sign $$\int {} $$ and $$\frac{d}{{dx}}$$ on the right side will cancel each other out, i.e.
\[\int {\frac{1}{x}dx} = \ln x + c\]

Example: Evaluate the integral $$\int {\frac{{1 – x}}{x}dx} $$ with respect to $$x$$

We have integral \[I = \int {\frac{{1 – x}}{x}dx} \]
\[\begin{gathered} \int {\frac{{1 – x}}{x}dx} = \int {\left( {\frac{1}{x} + \frac{x}{x}} \right)\,} dx \\ \Rightarrow \int {\frac{{1 – x}}{x}dx} = \int {\left( {\frac{1}{x} + 1} \right)\,} dx \\ \end{gathered} \]

Using integration of one over $$X$$, we have
\[\begin{gathered} \int {\frac{{1 – x}}{x}dx} = \int {\frac{1}{x}} \,dx + \int {1\,dx} \\ \Rightarrow \int {\frac{{1 – x}}{x}dx} = \ln x + x + c \\ \end{gathered} \]