Introduction to the integral of 1/x
In calculus, the integral is a fundamental concept that assigns numbers to functions to define displacement, area, volume, and all those functions that contain a combination of tiny elements. It is categorized into two parts, definite integral and indefinite integral. The process of integration calculates the integrals. This process is defined as finding an antiderivative of a function.
Integrals can handle almost all functions, such as trigonometric, algebraic, exponential, logarithmic, etc. This article will teach you what is integral to an algebraic function 1/x. You will also understand how to compute 1/x integral by using different integration techniques.
What is the integration of 1 by x?
The integration of 1 by x is an antiderivative of 1/x function which is equal to ln|x|. It is also known as the reverse derivative of the function 1/x which is an algebraic function. It can be calculated by using the power rule of integral. This rule is written as;
$\int x^n dx=\frac{x^{n+1}}{n+1}+c$
This formula says that the integral of any algebraic function with some exponent, can be calculated by adding 1 in its exponent and dividing by the new exponent i.e n+1.
1/x integration formula
The formula of integral of 1 by x contains integral sign, coefficient of integration and the function as 1/x. It is denoted by ∫(1/x)dx. In mathematical form, the integral of 1/x is:
$\int \frac{1}{x}=\ln x + c$
Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of integral.
How to find the integral 1/x?
The integration of 1/x is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate integral of x^-1 by using:
- Integration by parts
- Definite integral
Integral of 1 x by using integration by parts
The derivative of a function calculates the rate of change, and integration is the process of finding the antiderivative of a function. The integration by parts is a method of solving the integral of two functions combined together. Let’s discuss calculating the integral of 1 by x by using integration by parts.
Proof of integral of 1/(x) by using integration by parts
To integrate the function 1/x using integration by parts, we can use the following formula:
$I=f(x)\int g(x) dx - \int [f’(x)\int g(x) dx]dx$
Suppose that,
$f(x) = \frac{1}{x}$
$g(x) =1$
Using these values in the above formula,
$I=\frac{1}{x}\int 1dx -\int [\frac{d}{dx}(\frac{1}{x})\int 1dx)dx$
Integrating,
$I=\frac{1}{x}\times x -\int [\frac{-1}{x^2}\times x]dx$
$I=1+\int \frac{1}{x}dx$
$I=1+\ln x +c$
Hence the integration of 1/x by using integration by parts calculator is 1+ln x +c, where c is a constant known as an integration constant. But in actuality, the 1/x integral is equal to ln x +c.
Integral of x^-1 by using definite integral
The definite integral is a type of integral that calculates the area of a curve by using infinitesimal area elements between two points. The definite integral can be written as:
$∫^b_a f(x) dx = F(b) – F(a)$
Let’s understand the verification of the 1/x integration by using the definite integral.
Proof of integral of 1 x by using definite integral
To compute the integral of 1/x by using a definite integral, we can use the interval from a to b. It means that we can evaluate the definite integral of x square for any value of a and b. Let’s compute it from 1 to 2.
$\int^2_1 \frac{1}{x} dx=\ln x|^2_1 $
Substituting the values of upper and lower bounds.
$\int^2_1 \frac{1}{x} dx=\ln 2 -\ln 1$
Since the value of ln 1 is equal to zero.
$\int^2_1 \frac{1}{x} dx =\ln 2 =0.69315$
For any value of a and b, we can evaluate the definite integral by using the above formula. You can also use our definite integration calculator to make this calculation simple in just one click.
Frequently Asked Questions
What is the integration of x 4?
The antiderivative or the integral of x^4 is the same as the integral of x^2. It can be calculated as;
$\int x^4 dx = \frac{x^{4+1}}{4+1}=\frac{x^5}{5}+c$
How do you find the integral?
To find the integral, we use the fundamental theorem of calculus. This theorem states that if a function f is continuous on an interval [a, b], then,
∫abf(x) dx = F(b) – F(a)
Which is known as definite integral. Another formula to calculate the integral is the indefinite integral, that is;
∫f(x)=F(x) +c
What is the integral of 1/(1+x^2)?
The integral 1/1+x^2 is equal to tan^-1(x). Mathematically, it is expressed as;
$\int \frac{1}{1+x^2}dx =\tan^{-1}x +c$
Where c is the constant of integration.
