Question

What is the integration of $\dfrac{1}{x}$ ?

Answer 1

Hint: We need to find out the integral of the function $\dfrac{1}{x}$ . Firstly, we start to solve the problem by finding the derivative of $\dfrac{1}{x}$ using the fundamental theorem of calculus and chain rule. Then, we find the antiderivative of $\dfrac{1}{x}$ to get the desired result.

Complete step by step answer:
We are given a function and need to integrate the function. We will be solving the given question using the concepts of fundamental theorem and chain rule in calculus.
We have differentiation and integration as a part of calculus. The derivative of a particular function is considered to be the rate of change in function due to change in one of its variables. Whereas we have integration, in calculus, which is a method to find the values of definite and indefinite integrals.
The fundamental theorem of calculus links the two major concepts in calculus namely, Differentiation and Integration.
We know that the first fundamental theorem of calculus states that for any continuous function $f$ and $F$ is the indefinite integral on $\left[ p,q \right]$ then,
$\Rightarrow \int\limits_{p}^{q}{f\left( x \right)dx=F\left( q \right)-F\left( p \right)}$
We also know that the first fundamental theorem of calculus states that for any continuous function $f$ , $F$ is defined as,
$\Rightarrow F\left( x \right)=\int\limits_{p}^{x}{f\left( t \right)dt}$
From the fundamental theorem of calculus, we get,
$\Rightarrow \dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$ for the value of $x>0$
The chain rule is used to find the derivatives of the composite functions. It is usually given as follows,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}$
Here,
$\dfrac{dy}{dx}$ is the derivative of y with respect to x
$\dfrac{dy}{dt}$ is the derivative of y with respect to t
$\dfrac{dt}{dx}$ is the derivative of t with respect to x
Applying chain rule to the above expression, we get,
$\Rightarrow \dfrac{d}{dx}\left( \ln \left( -x \right) \right)=\dfrac{1}{x}$ for the value of $x<0$
Integrating the above equations on both sides, we get,
$\Rightarrow \int{\dfrac{1}{x}dx=\ln \left( x \right)}+c$ for the value of $x>0$
$\Rightarrow \int{\dfrac{1}{x}dx=\ln \left( -x \right)+c}$ for the value of $x<0$
The value of $\ln \left( x \right)$ and $\ln \left( -x \right)$ can be written as $\ln \left| x \right|$
Substituting the same, we get,
$\therefore \int{\dfrac{1}{x}dx=\ln \left| x \right|+c}$

Note: One of the common mistakes we make is assuming that the functions $\log x$ and $\ln x$ are the same. $\log x$ means the base 10 logarithmic function and $\ln x$ is the base e logarithmic function. The integral of $\dfrac{1}{x}$ is $\ln \left| x \right|+c$ and not $\log \left| x \right|+c$ .