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}