## Introduction to the integral of 1/(1+x^2)

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 1/1+x^2. You will also understand how to compute 1/1+x^2 integral by using different integration techniques.

## What is the integral of 1 by 1+x^2?

The integral of 1/x^2+1 is an antiderivative of the 1/1+x^2 function which is equal to ½ ln(1-x/1+x). It is also known as the reverse derivative of the function 1/1+x^2 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.

### Integral of 1/(1+x^2) formula

The formula of integral of square root x squared contains integral sign, coefficient of integration and the function as 1/1+x^2. It is denoted by ∫1/1+x^2 dx. In mathematical form, the integral of 1/1+x^2 is:

$$\int \frac{1}{1+x^2}=\tan^{-1}x+c$$

Where c is any constant involved, dx is the coefficient of integration calculator and ∫ is the symbol of integral.

## How do you integrate 1/1+x^2?

The integral of 1/1+x^2 is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate integral of 1 by 1+x^2 by using:

- Trigonometric substitution method
- Integration by partial fraction

## Integral of 1/1+x^2 by using trigonometric substitution

The trigonometric substitution is a method of integration in calculus. It is used to calculate the integral of a function which is complex to be calculated by usual integration. Let’s discuss calculating the integral of the 1/1+x^2 by using the trig-substitution calculator.

### Proof of Integral of square root of x by using substitution method

To proof the integral of 1/1+x^2 by using substitution method, suppose that:

$$I=\int \frac{1}{1+x^2}dx $$

In the trigonometric substitution, we substitute the variable with a trigonometric function. Now suppose that,

$$x=\tan \theta $$

and

$$dx=\sec^2 \theta d\theta $$

Using the new substitution in the above integral.

$$I=\int \frac{\sec^2 \theta}{1+\tan^2\theta}d\theta $$

Since,

$$\sec^2 \theta =1+\tan^2 \theta$$

We can write it as;

$$I=\int \frac{\sec^2 \theta}{\sec^2 theta}d\theta =\int 1d\theta $$

Integrating,

$$I=\theta $$

Now substituting the value of ,

$$I=\tan^{-1}\theta +c$$

Hence we have verified the 1/1+x^2 integral by using the trigonometric substitution method. The u-substitution method is also another way to evaluate integrals.

**Integral of 1/1+x^2 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 partial fraction is a method of solving the integral by breaking down the functions into different parts. Let’s discuss calculating the integral of 1/1+x^2 by using integration by partial fraction.

**Proof of integral of 1 by (1+x^2) by using integration by parts**

To integrate the function 1/1+x^2 using integration by parts, we can use the following formula:

I=f(x)g(x)dx-[ddx(f(x)g(x)dx]dx

Suppose that,

$$f(x)=1$$

$$g(x)=\frac{1}{1+x^2}$$

Using these values in the above formula,

$$I=1\int \frac{1}{1+x^2} dx-\int [\frac{d}{dx}(1)\int \frac{1}{1+x^2}dx]dx $$

Integrating,

$$I=\tan^{-1}x - \int (0)\tan^{-1}x dx $$

Simplifying,

$$I=\tan^{-1}x+c$$

Hence the integral of 1 by 1+ x^2 by using by parts integration 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.

## Frequently Asked Questions

### What are the 5 basic integration formulas?

5 basic formulas of integration are:

- Power rule formula

$$\int x^n dx=\frac{x^{n+1}}{n+a}$$

- Integration by Parts

$$\int f(x)g(x)dx = f(x)\int g(x)dx - \int(f’(x)\int g(x)dx)dx$$

- Difference rule

$$\int[f(x)-g(x)] = \int f(x) dx - \int g(x)dx$$

- Sum rule

$$\int[f(x)+g(x)] = \int f(x) dx + \int g(x)dx$$

- Multiplication by Constant

$$\int af(x)dx=a\int f(x)dx$$

### What is the integral of 1 by 1+x^2?

The integral of 1/ 1+x2 is the reverse derivative of 1/1+x2 which is equal to the inverse of tan. Mathematically, it can be written as;

$$\int \frac{1}{1+x^2}dx = \tan^{-1}x + c$$