Integration of One over X

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 , so the integral sign \int  {} and \frac{d}{{dx}} on the right side will cancel each other, 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}

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