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.

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