Let $x$ be a variable, which represents a non-zero real number. The multiplicative inverse or reciprocal of variable is written as $\dfrac{1}{x}$ mathematically. The indefinite integral of $\dfrac{1}{x}$ with respect to $x$ is written in integral calculus in the following mathematical form.

$\displaystyle \int{\dfrac{1}{x} \,}dx$

According to differentiation, the derivative of natural logarithm of $x$ with respect to $x$ is equal to multiplicative inverse or reciprocal of $x$. So, it is written in the following mathematical form.

$\dfrac{d}{dx}{\, \log_e{x}} \,=\, \dfrac{1}{x}$

$\implies$ $\dfrac{d}{dx}{\ln{x}} \,=\, \dfrac{1}{x}$

The derivative of a constant is zero as per differential calculus. So, it does not affect the differentiation when an arbitrary constant is included in differentiation by adding it to natural logarithmic function.

$\implies$ $\dfrac{d}{dx}{(\ln{x}+c)} \,=\, \dfrac{1}{x}$

According to integral calculus, the collection of all primitives of multiplicative inverse function $\dfrac{1}{x}$ is called the indefinite integral of $\dfrac{1}{x}$ function with respect to $x$ and it is written in the following mathematical form in mathematics.

$\displaystyle \int{\dfrac{1}{x} \,}dx$

Actually, the primitive or an antiderivative of $\dfrac{1}{x}$ is $\log_e{x}$ and the arbitrary constant becomes the constant of integration.

$\dfrac{d}{dx}{(\ln{x}+c)} = \dfrac{1}{x}$ $\,\Longleftrightarrow\,$ $\displaystyle \int{\dfrac{1}{x} \,}dx = \log_e{x}+c$

$\therefore \,\,\,\,\,\,$ $\displaystyle \int{\dfrac{1}{x} \,}dx = \ln{x}+c$

Therefore, it has proved that the anti-derivative or indefinite integration of multiplicative inverse or reciprocal function is equal to the sum of the natural logarithm and the constant of integration.

Latest Math Topics

Dec 13, 2023

Jul 20, 2023

Jun 26, 2023

Latest Math Problems

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved