[Maths Class Notes] on Indefinite Integral Formula Pdf for Exam

In calculus, integration is just the reverse of differentiation. Integration is a process with which we can find a function with its given derivative. The integration indefinite formulas can be applied to different functions. 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 the 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 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.

 

Some Properties of Indefinite Integral

1. [int cf(x)dx = cint f(x) dx]

 

Here, c = constant value

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

 

2. [int -f(x)dx = cint f(x) dx] 

 

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

 

3. [int f(x) pm g(x) dx = 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. 

 

Indefinite Integral Formulas

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.

 

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

2. [int e^x dx = e^{n + 1} + C ]

3. [int frac{1}{x} dx = ln x + C ]

4. [int sinx dx = -cos x+ C ]

5. [int cosx dx = sinx + C ]

6. [int sec^2x dx = tanx + C]

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

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

9. [int log_ax dx = frac{1}{ln a} . frac{1}{x} + C]

10. [int frac{1}{ sqrt{1 – x^2}}  dx= arc sinx + C]

11. [int frac{1}{ sqrt[x]{x^2 – 1}}  dx= arc sinx + C]

 

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

 

Solved Examples for Indefinite Integral Formulas

1. Use the indefinite integral formulas to evaluate the following integral.

 

[int 6x^5 – 2x + frac{3}{x} dx]

 

Solution: 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:

 

[int 6 frac{x^6}{6} – 2frac{x^2}{2} + frac{3}{x} ln x + C dx]

 

[x^6  – x^2 + 3 ln x + C]

 

Here C is represented as an integral constant.

 

2. Evaluate the given indefinite integral.

 

[int x^4 + 6x – 10dx]

 

Solution: Given indefinite integral is

 

[int x^4 + 6x – 10dx]

 

Therefore,

 

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

 

[= frac{1}{5} x^5 + 3 x^2 – 10x + C]

 

Here, C represents the integral constant.

 

3. Evaluate the given indefinite integral.

 

[int (3x^2 – 6x + 2cosx) dx]

 

Solution We will apply the properties 1 and 2

 

[I = int (3x^2 – 6x + 2cosx) dx]

 

= [int 3x^2 dx – int 6x dx + int 2cosx dx]

 

= [3 int x^2 dx – 6 int x dx + 2 int cosx dx]

 

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

 

[I = 3. frac{x^3}{3}  – 6. frac{x^2}{2} + 2. Sinx + C = x^{3} – 3x^{2} + 2sinx + C.]

 

4. Find the indefinite integral.

 

[int (1 + x)(1 + 2x) dx]

Solution: 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 3xdx + int 1 dx]

 

[= 2 int x^2 dx + 3 int xdx + int dx = 2. frac{x^3}{3} + 3.frac{x^2}{2} + x]

 

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

 

5. Find the indefinite integral.

 

[int begin{pmatrix} frac{1}{x^2} – frac{1}{x^3} end{pmatrix} dx]

 

Solution: According to the sum rule,

 

[I = int begin{pmatrix} frac{1}{x^2} – frac{1}{x^3} end{pmatrix} 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 dx = – frac{1}{x} + frac {1}{2x^{2}} + C ]

 

6. Calculate

 

[int (sqrt{x} + sqrt[n]{x}) dx]

 

Solution 

 

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

 

7. Find the indefinite integral

 

[int frac{x + 1}{sqrt{x}} dx]

 

Solution

 

[int frac{x + 1}{sqrt{x}} dx = int begin{pmatrix} frac{x}{sqrt{x}} + frac{1}{sqrt{x}} end{pmatrix} = int begin{pmatrix} sqrt{x} + frac{1}{sqrt{x}} end{pmatrix} = int sqrt{x} dx + int frac{1}{sqrt{x}} ]

 

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

Points To Remember

• Integration or integrating f(x) is also the process of finding the indefinite integral of a function. 

• Suitable formulas are used in getting the antiderivative of the function.

• A family of functions is represented by an indefinite integral and derivatives of the function are f.

• The indefinite integral is quite similar to the definite integral but both are not the same as indefinite integral results in a function, while definite integral results in a real number.

• Only the indefinite integrals contain an actual number C in the process of integration.

• Indefinite integral contains no bounds, while a definite integral contains the upper and lower limits, that is, the start and end value.

• The integral of 0 is C or any constant because the derivative of any constant is zero and hence, ∫0 dx = C.

• An indefinite integral is a function that will explain an area under the function’s curve from an undefined point to another arbitrary point.

• There are five methods of integration namely, the substitution method, integration by partial fractions, solve integration is by parts, Euler substitution, and the reduction method.

• Integral Calculus is particularly used for

1. For calculating f from f’.

2. If a function f is differentiable in the interval of consideration, then f’ is defined. 

3. In differential calculus, it has been already known to calculate derivatives of a function, but it can be undone with the help of integral calculus.

4. It is used for calculating the area under a curve.