# 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
$\begin{gathered} \int {\frac{{1 – x}}{x}dx} = \int {\frac{1}{x}} \,dx + \int {1\,dx} \\ \Rightarrow \int {\frac{{1 – x}}{x}dx} = \ln x + x + c \\ \end{gathered}$