Evaluate the following definite integral: $\int\limits_{2}^{3}{\dfrac{1}{x}dx}$.

Hint: We start solving the problem by finding the indefinite part of the integral by using $\int{\dfrac{1}{x}={{\log }_{e}}x+C}$. We then substitute the limits in the obtained result of indefinite integral. We make necessary calculations for the value obtained to get the required value.

Complete step by step answer:
According to the problem, we are given that to find the value of the definite integral $\int\limits_{2}^{3}{\dfrac{1}{x}dx}$.
We first find the value of indefinite integral $\int{\dfrac{1}{x}dx}$ and then substitute the results.
We have got the value of $\int{\dfrac{1}{x}dx}$ ---(1).
We know that the integration of the function $\dfrac{1}{x}$ is defined as $\int{\dfrac{1}{x}={{\log }_{e}}x+C}$. We use this result in equation (1).
We have got the value of $\int{\dfrac{1}{x}dx}={{\log }_{e}}x$ ---(2). (Here adding arbitrary constant is neglected as it cancels out after substituting the limits to the function obtained after integration).
We now substitute the limits of the integral in the function obtained in equation (2).
We know that the definite integral is defined as $\int\limits_{a}^{b}{f'\left( x \right)dx=f\left( b \right)-f\left( a \right)}$. We use this result in equation (2).
So, we have got the value of $\int\limits_{2}^{3}{\dfrac{1}{x}dx}=\left. {{\log }_{e}}x \right|_{2}^{3}$.
We have got the value of $\int\limits_{2}^{3}{\dfrac{1}{x}dx}=\left( {{\log }_{e}}\left( 3 \right)-{{\log }_{e}}\left( 2 \right) \right)$ ---(3).
We know that the subtraction of two logarithmic functions is defined as ${{\log }_{e}}\left( a \right)-{{\log }_{e}}\left( b \right)={{\log }_{e}}\left( \dfrac{a}{b} \right)$. We use this result in equation (3).
We have got the value of $\int\limits_{2}^{3}{\dfrac{1}{x}dx}={{\log }_{e}}\left( \dfrac{3}{2} \right)$.
We have got the value of $\int\limits_{2}^{3}{\dfrac{1}{x}dx}={{\log }_{e}}\left( 1.5 \right)$.
We have found the value of the definite integral $\int\limits_{2}^{3}{\dfrac{1}{x}dx}$ as ${{\log }_{e}}\left( 1.5 \right)$.
∴ The value of the definite integral $\int\limits_{2}^{3}{\dfrac{1}{x}dx}$ is ${{\log }_{e}}\left( 1.5 \right)$.

Note:
We need not add constant function while we are solving problems containing definite integrals. We should not apply $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c}$ by taking the value of n as ‘–1’, as this formula is not valid for n = –1. If the power of x is less than –1 in the denominator, we can use that formula.