- IIT JEE Study Material

## Indefinite Integrals Problems and Solutions

In calculus, integration is defined as the inverse process of differentiation, and hence the evaluation of an integral is called an antiderivative. It is a process of the summation of a product if we have not said the summation is to be done from which point to which point. In other words, the interval of summation is indefinite, and hence these types of integrals are known as indefinite integrals . Integration is an important part of the calculus, which includes single integral, double integral, and multiple integrals. Various types of integrals are used to find the surface area and the volume of geometric solids.

## What Is Infinite Integral in Calculus?

The indefinite integral is defined as a function that describes an area under the function’s curve from an undefined point to another arbitrary point. The absence of a specified first point leads to an arbitrary constant, most often denoted as C, and is always considered a part of an indefinite integral.

In other words, the indefinite integral is the family of all functions whose derivative is f(x) but with a possibly finite set of exceptions. \(\begin{array}{l}\frac{d}{dx}\int f(x)dx = f(x)\end{array} \) .

Indefinite integrals are integrals without limits. The technique of integration is very useful in two ways.

- To find the function whose derivative is given.
- To find the area bounded by a curve given by the function under certain conditions.

## Rules of Integration

The antiderivative of a definite integral is only implicit, which means the solution will only be in a functional form. That is, ∫f(x)dx = g(x) + C, where g(x) is another function of x and C is an arbitrary constant. However, there are many integrals which are to be integrated within a given interval. \(\begin{array}{l}\text{They are denoted in general as}\ \int_a^b f(x) dx\end{array} \) , where a and b are the limits of the interval. Such types of integrals are known as definite integrals. \(\begin{array}{l}\text{The solution of a definite integral is unique and the solution to}\ \int_a^b f(x)dx = F(b) – F(a),\end{array} \) , where F(x) is the antiderivative of the given integral.

There are a few important rules for integration:

Integration of some functions may be readily done for functions whose derivatives are known. But, in many cases, the integrant (the function to be integrated) may not be that simple. It may be a sum, difference, product or quotient of two functions. To perform the integration of such functions, we need to follow the basic integration rules. Some basic integration rules are given below.

Rule 1: The integration of a sum or difference of two functions is the sum or difference (respectively) of the integration of the individual functions.

∫[f(x) + g(x)]dx = ∫f(x)dx + ∫g(x)dx

Rule 2: The integration of a product of a function is [ First function * the integration of the second function – integration of { the integration of the second function*derivative of the first function}].

If u and v are two functions, ∫udv = uv − ∫vdu.

To memorise better, we chose dv as the second function. Hence, in all the integration of the product of two functions, we equate the second function as dv and immediately figure out what is v. Also, you should know the technique of correctly assigning the first and second functions.

There are many ways to find the integration of a given function, such as:

- Integration by Parts
- Integration by Substitution Method or Change of Variable
- Directly Using the Formula
- Integration by Partial Fraction Method

## Solved Problems on Indefinite Integrals for JEE

Practice the below problems to crack your exam.

Question 1: Solve ∫(x 2 + 3x – 2)dx

Solution: ∫(x 2 + 3x – 2)dx = (x 3 /3)+ (3x 2 )/2 – 2x + c.

Question 2: Solve ∫ 4 x e 2x dx

Solution: ∫ 4 x e 2x dx = ∫ 2 2x e 2x dx = ∫ (2e) 2x dx =(2e) 2x /(2 +log 4) + c

= tan(3x)/6 + C

Let us use the substitution method to solve the problem.

Put x 5 = t, then

Differentiating it, we have

⇒ 5x 4 dx = dt

⇒ x 4 dx = dt/5

Putting this value in equation (1), we have

⇒ (1/5) tan -1 x 5 + C

Let tan -1 m 3 = t

⇒ 3m 2 /(1 + m 6 ) dm = dt

⇒ m 2 (1 + m 6 ) dm = dt/3

Substituting this value in equation (1), we have

= t 2 /2(3) + C

Let cos 2 4x = t

⇒ – 2cos(4x).sin(4x) dx = dt

⇒ – sin(8x) dx = dt

⇒ sin(8x) dx = -dt

= -sin -1 t + c

= -sin -1 cos 2 4x + c.

⇒ D’(3x 2 +6x+2) = 6x + 6

So, 6x + 8 = A(6x + 6) + B

By comparing coefficients, we have

⇒ A = 1 and B = 2

So, we have

= I 1 + I 2 …(1)

Where, \(\begin{array}{l}I_{1} = \int \frac{6x+6}{3x^2+6x+2}dx….(2)\end{array} \) and

Solve for I 1

Let 3x 2 + 6x + 2 = t

⇒ (6x + 6) dx = dt

Substituting this value in equation (2), we have

Question 9: Solve ∫ e 2x cos(3x + 4) dx

Let I = ∫ e 2x cos(3x+4) dx

Taking e 2x as the first and cos(3x + 4) as the second function and applying integration by parts, we have

⇒ I = ∫ e 2x cos(3x+4) dx

⇒ I = e 2x ∫ cos(3x+4) dx – ∫[D'(e 2x )∫ cos(3x+4)dx]dx

Let I = ∫[sin(ln y) + cos(ln y)] dy….(1)

Let, ln y = t

⇒ dy = e t dt

Substituting in equation (1), we have

⇒ I = ∫ e t [sin(t) + cos(t)] dt

D’(sin(t)) = cos(t)

Using the result ∫ e x [f(x) + f’(x)] dx = e x f(x) + c, we can say

⇒ I=∫ e t [sin(t) + cos(t)] dt = e t sin(t) + c

⇒ I = e (ln y) sin(ln y) + c

⇒ I = y. sin(ln y) + c .

## Indefinite Integration – Video Lesson

## Indefinite-Integration Problems

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## Integral Calculus

In these lessons, we introduce a notation for antiderivatives called the Indefinite Integral. We also give a list of integration formulas that would be useful to know.

Related Pages Calculus: Integration Calculus: Derivatives Calculus Lessons

Indefinite Integrals

The following is a table of formulas of the commonly used Indefinite Integrals. You can verify any of the formulas by differentiating the function on the right side and obtaining the integrand. Scroll down the page if you need more examples and step by step solutions of indefinite integrals.

Table Of Indefinite Integral Formulas

Example: Find the general indefinite integral.

## Definite Integrals And Indefinite Integrals

The connection between the definite integral and indefinite integral is given by the second part of the Fundamental Theorem of Calculus .

If f is continuous on [a, b] then

Definition Of Indefinite Integrals

An indefinite integral is a function that takes the antiderivative of another function. It is visually represented as an integral symbol, a function, and then a dx at the end. The indefinite integral is an easier way to symbolize taking the antiderivative. The indefinite integral is related to the definite integral, but the two are not the same.

Antiderivatives And Indefinite Integrals

Example: What is 2x the derivative of? This is the same as getting the antiderivative of 2x or the indefinite integral of 2x.

Indefinite integrals are functions that do the opposite of what derivatives do. They represent taking the antiderivatives of functions.

A formula useful for solving indefinite integrals is that the integral of x to the nth power is one divided by n+1 times x to the n+1 power, all plus a constant term.

Indefinite Integrals, Step By Step Examples

Step 1: Add one to the exponent Step 2: Divide by the same. Step 3: Add C

Example: ∫3x 5 , dx

More Indefinite Integral, Step By Step, Examples: With Square Root

Example: ∫3√x, dx

More Indefinite Integral, Step By Step, Examples: x In The Denominator

Example: ∫6/x 4 , dx

Complicated Indefinite Integrals

Not all indefinite integrals follow one simple rule. Some are slightly more complicated, but they can be made easier by remembering the derivatives they came from. These complicated indefinite integrals include the integral of a constant (the constant times x), the integral of e x (e x ) and the integral of x -1 (ln[x]).

Indefinite Integration (Polynomial, Exponential, Quotient)

How to determine antiderivatives using integration formulas?

- ∫(3x 2 - 2x + 1) dx

Basic Integration Formulas

Here are some basic integration formulas you should know.

## Definite Integral

The Definite Integral - Understanding the Definition.

Calculating A Definite Integral Using Riemann Sums - Part 1

This video shows how to set up a definite integral using Riemann Sums. The Riemann Sums will be computed in Part 2.

Calculating A Definite Integral Using Riemann Sums - Part 2

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