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In calculus, integration is just the reverse of differentiation. Integration is a process with which we can find a function with its given derivative. These integration indefinite formula can be applied to different functions. This integration can either be indefinite or definite type. In an algebraic method, Integration is the way to understand the concept of indefinite integral and find the integral for some mathematical function at any point. The integral that comes after the process helps to determine the function from its derivatives. Additionally, the concept of indefinite integral is also useful in solving many problems in mathematics as well as science.

The indefinite integrals are not bounded but free from both the endpoints. It means that the independent variable will not carry any given interval. For definite integration, both endpoints are quite specific and definite whereas, for the indefinite integrals, there are no boundaries.

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Thus we may find an equation instead of having a set of boundary values to produce the integral due to differentiation with no boundaries. We need not use the values to get a definite answer.

For a given, f(x), F’(x) will be considered as its derivative. It is obvious that the most general antiderivative of the function f(x) will be an indefinite integral.

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

\[\int f(x) dx\]

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

\[\frac{d}{dx} f(x) = g(x)\]

Thus,

\[g(x) dx = f(x) + C\]

Here, C represents the constant of integration which is a must. For computation, definite integrals always represent some bounded region. This is not the same with indefinite integral as they are the bounded region of the curve.

1) \[\int c f(x) dx = c \int f(x) dx\]

Here, c = constant value

From the indefinite integral, we can take out the multiplicative constants.

2) \[\int -f(x) dx = \int f(x) dx\]

Due to the negative function, the indefinite integral is also negative.

3) \[\int f(x) \pm g(x) d = \int f(x) dx \pm \int g(x) dx\]

It shows the sum as well as the difference of the integral of the functions as the sum or the difference of their individual integral. So, these were some properties of indefinite integral.

Given below are the important indefinite integral formulas. Also, there are solved examples for indefinite integral formulas that you can practice after going through the indefinite formula.

\[\int x^{n} dx = \frac{1}{n + 1} x^{n + 1} + C\] unless n = -1

\[\int e^{x} dx = e^{x} + C\]

\[\int \frac{1}{x} dx = \text{ln }x + C\]

\[\int sin x dx = -cos x + C\]

\[\int cos x dx = sinx + C\]

\[\int sec^{2}x dx = tanx + C\]

\[\int \frac{1}{1 + x^{2}} dx = \text{arc tan x} + C\]

\[\int a^{x} dx = \frac{a^{x}}{ln a} + C\]

\[\int log_{a} x dx = \frac{1}{ln a} \cdot \frac{1}{x} + C\]

\[\int \frac{1}{\sqrt{1 - x^{2}}} dx = \text{arc sin x} + C\]

\[\int \frac{1}{\sqrt[x]{x^{2} - 1}} dx = \text{arc sec x} + C\]

So these were all the indefinite formulas you need to know.

Example 1) Use the indefinite integral formulas to evaluate the following:

\[\int 6x^{5} - 2x + \frac{3}{x} dx\]

Solution 1) The integral that we are provided with is

\[\int 6x^{5} - 2x + \frac{3}{x} dx\]

So the suitable indefinite integral formulas that we can get is:

\[\frac{6x^{2}}{6} - \frac{2x^{2}}{2} + 3 ln x + C\]

\[x^{6} - x^{2} + 3 ln x + C\]

Here C is represented as integral constant.

Example 2) Evaluate the given indefinite integral

\[\int x^{4} + 6x -10 dx\]

Solution 2) Given indefinite integral is

\[\int x^{4} + 6x -10 dx\]

Therefore,

= \[\frac{1}{5} x^{5} + \frac{6}{2} x^{2} - 10 x + C\]

= \[\frac{1}{5} x^{5} + 3x^{2} - 10 x + C\]

Here, C represents the integral constant.

Example 3) Evaluate the given indefinite integral

\[\int (3x^{2} - 6x + 2 cos x)dx\]

Solution 3) We will apply the properties 1 and 2

\[I = \int (3x^{2} - 6x + 2 cos x) dx = \int 3x^{2} dx - \int 2 cos x dx = 3 \int x^{2} dx - 6 \int x dx + 2 \int cos x dx\]

Now using the integral tables, we can evaluate all the three integral

\[I = 3 \cdot \frac{x^{3}}{3} - 6 \cdot \frac{x^{2}}{2} + 2 \cdot sin x + C = x^{3} - 3 x^{2} + 2 sinx + C.\]

Example 4) Find the indefinite integral \[\int (1 + x) (1 + 2x) dx\]

Solution 4) First we have to simplify the integrand:

\[(1 + x)(1 + 2x) = 1 + x + 2x + 2x^{2} = 2x^{2} + 3x + 1\]

The integral given:

\[\int (1 + x)(1 + 2x) dx = \int (2x^{2} + 3x + 1)dx = \int 2x^{2} dx + \int 3x dx + \int 1 dx = 2 \int x^{2} dx + 3 \int xdx + \int dx = 2 \cdot \frac{x^{3}}{3} + 3 \cdot \frac{x^{2}}{2} + x\]

= \[\frac{2x^{3}}{3} + \frac{3x^{2}}{2} + x + C\]

Example 5) Find the indefinite integral \[\int (\frac{1}{x^{2}}- \frac{1}{x^{3}} dx\]

Solution 5) According to the sum rule,

\[I = \int (\frac{1}{x^{2}} - \frac{1}{x^{3}} dx = \int \frac{dx}{x^{2}} - \int \frac{dx}{x^{3}}\]

In both the integrals, the integrands are the power functions, so

\[I = \int x^{-2} dx - \int x^{-3} dx = \frac{x^{-1}}{-1} - \frac{x^{-2}}{-2} + C = -\frac{1}{x} + \frac{1}{2x^{2}} + C\]

Example 6) Calculate \[\int (\sqrt{x} + \sqrt[3]{x})dx\]

Solution 6) \[\int (\sqrt{x} + \sqrt[3]{x})dx = \int \sqrt{x} dx + \int \sqrt[3]{x}dx = \int x^{\frac{1}{2}} dx + \int x^{\frac{1}{3}} dx\]

= \[\frac{x^{\frac{3}{2}}}{\frac{3}{2}} + 2\sqrt{x} + C = \frac{2\sqrt{x^{3}}}{3} + 2 \sqrt{x} + C\]

Example 7) Find the indefinite integral \[\int \frac{x + 1}{\sqrt{x}} dx\]

Solution 7) \[\int \frac{x + 1}{\sqrt{x}} dx = \int (\frac{x}{\sqrt{x}} + \frac{1}{\sqrt{x}})dx = \int (\sqrt{x} + \frac{1}{\sqrt{x}})dx = \int \sqrt{x}dx + \int \frac{dx}{\sqrt{x}}\]

= \[\frac{x^{\frac{3}{2}}}{\frac{3}{2}} + 2\sqrt{x} + C = \frac{2\sqrt{x^{3}}}{3} + 2 \sqrt{x} + C\]

FAQ (Frequently Asked Questions)

Question 1) How can we Memorize the Indefinite Integral Formulas?

Answer 1) We all face a situation where we memorize all the important formulas but exactly at the crucial moment when we need them, we tend to forget all the formulas. This happens especially at the time of examination. But hold! We have a solution. There are various methods we can practice in order to remember what we memorized. There might be one or two methods that may not work for you so see what works for you and choose wisely.

**Method 1)** Make a list of all the formulas and keep it in front of you so that you can see and go through them every day.

**Method 2)** Practice… Practise….. Practice every day. Practice proving formulas as much as you can.

**Method 3)** Don’t just mug up and practice it blindly but try to understand the logic behind it.

**Method 4)** Sing out the formulas and record it. Hear it every day.

**Note:** Don’t just do it for your upcoming exams but prepare for the long run.

Question 2) What are Derivatives?

Answer 2) Derivation is the rate of change of a function concerning the independent variable. We use derivatives when the rate of change is not constant. It is also used to measure the sensitivity of one variable with respect to the other variable. The process that we use to find the derivatives is called derivation. Its inverse process is called anti-derivation. Derivatives can be first-order derivative or second-order derivatives. The first-order derivative helps us to know the direction of the function and the second-order derivative tells us about the shape of the graph of the function.