In calculus, integration is a reverse process of differentiation. Integration is the process to find a function with its given derivative. These integration formulas are applicable to different functions. This integration may be indefinite or definite type. This article will explain the concept of indefinite integral with indefinite integral formulas and examples. Let us learn the concept!

**Concept of Indefinite Integral**

Integration is an algebraic method to find the integral for some mathematical function at any point. The integral comes after the process will help to determine the function from its derivatives. It also helps to solve many problems in mathematics as well as science.

The indefinite integrals are not bounded from both the endpoints. It means the independent variable will not have any given interval. But for definite integration, both endpoints are specific and definite.

Source: en.wikipedia.org

In the indefinite integrals, there will not exist any boundaries. Thus instead of having a set of boundary values, we may find an equation to produce the integral due to differentiation with no boundaries. It will not need to use the values to get a definite answer.

For a given, f(x), Fâ€™(x) will be its derivative. Obviously the most general anti-derivative of the function f(x) will be indefinite integral.

So, the integral of a function f(x) with respect to variable x is:

\(\int f(x)\;dx\)

Also, integration is the inverse operation to the differentiation means that if,

\({d\over dx}f(x)=g(x)\\\)

Thus \(\int g(x)\;dx=f(x)+C . \\\)

The term C is known as the constant of integration, and it is must. Definite integrals are always representing some bounded region, for the computation. But, indefinite integrals are the unbounded region of the curve.

## Some Properties ofÂ Indefinite Integral

- \(\displaystyle \int{{c\,f\left( x \right)\,dx}} = c\int{{f\left( x \right)\,dx}}\) Where c is any constant value. So, we take out the multiplicative constants from the indefinite integrals.

- \(\displaystyle \int{{ – f\left( x \right)\,dx}} = – \int{{f\left( x \right)\,dx}}\) This is showing negative indefinite integrals due to negative function.

- \(\displaystyle \int{{f\left( x \right) \pm g\left( x \right)\,dx}} = \int{{f\left( x \right)\,dx}} \pm \int{{g\left( x \right)\,dx}}\) It shows the sum and difference of the integrals of functions as the sum or difference of their individual integrals.

**Formula for Indefinite Integrals**

- \(\int x^n\; dx = {1\over n+1}x^{n+1}+C \\ , \hbox{ unless n=-1 } \\\)
- \(\int e^x \;dx= e^x+C \\\)
- \(\int {1\over x} \;dx= \ln x+C \\\)
- \(\int \sin x\;dx=-\cos x+C \\\)
- \(\int \cos x\;dx= \sin x + C\\\)
- \(\int \sec^2 x\;dx=\tan x+C \\\)
- \(\int {1\over 1+x^2} \; d=\arctan x+C\)
- \(\int a^x \;dx= {a^x\over \ln a}+C \\\)
- \(\int \log_a x\;dx={1\over \ln a}\cdot{1\over x}+C \\\)
- \(\int { 1 \over \sqrt{1-x^2 }} \; dx=\arcsin x+C\\\)
- \(\int { 1 \over x\sqrt{x^2-1 }} \; dx=\hbox{ arcsec}\, x+C\)

## Solved Examples forÂ Indefinite Integral Formulas

Q.1: Evaluate the following using Indefinite Integral Formulas: \(\int 6x^5-2x+{3\over x}\;dx\\\)

Solution: Given integral is: \(\int (6x^5-2x +{3\over x})\;dx\\\)

Applying the suitable formula we will get:

\({6x^6\over 6}-{2x^2\over 2}+3\ln x+C\\\)

\(x^{6}-x^{2}+3 \ln x + C\)

Where C is integral constant.

Q.2: Evaluate the following indefinite integral: \(\int{{{x^4} + 6x – 10\,dx}}\\\)

Solution: We need to evaluate the indefinite integral. Therefore we may apply suitable formula as below:

The indefinite integral is, \(\int{{{x^4} + 6x – 10\,dx}} \\\)

\(= \frac{1}{5}{x^5} + \frac{6}{2}{x^2} – 10x + C\\\)

\(= \frac{1}{5}{x^5} + 3 {x^2} – 10 x + C\)

Where c is integral constant.

I get a different answer for first example.

I got Q1 as 20.5

median 23 and

Q3 26