# Integration Formulas: Examples and Solutions

Integration is a fundamental concept in calculus that allows us to find the area under a curve or the accumulation of a quantity over an interval. The Integration Formula is a set of mathematical expressions that help us evaluate integrals efficiently. Simply put, Integration is like finding the total amount of something by adding up tiny pieces of them over a range. Similar to calculating the entire area under a curve by summing countless tiny rectangles that fit beneath it. Read on to learn more about the different integration formulas and their applications.

## Integration Formula by Parts

The Integration by Parts Formula is used to evaluate integrals where the integrand is a product of two functions, one of which is easy to differentiate, and the other is easy to integrate.

The Formula is:

∫ u dv = uv – ∫ v du

Where:

• u and dv are the first function and its differential, respectively.
• v and du are the second function and its differential, respectively.

Example 1:

Evaluate the integral ∫ x e^x dx

Solution:

Let u = x and dv = e^x dx

Then, du = dx and v = e^x

Applying the Integration by Parts Formula:

∫ x e^x dx = x e^x – ∫ e^x dx

∫ x e^x dx = x e^x – e^x + C

Example 2:

Evaluate the integral ∫ ln(x) dx

Solution:

Let u = ln(x) and dv = dx

Then, du = 1/x dx and v = x

Applying the Integration by Parts Formula:

∫ ln(x) dx = x ln(x) – ∫ x (1/x) dx

∫ ln(x) dx = x ln(x) – x + C

## Integration Formula by Substitution

The Integration by Substitution Formula, also known as the change of variable formula, is used to evaluate integrals where the integrand can be expressed in terms of a new variable.

The Formula is:

∫ f(g(x)) g'(x) dx = ∫ f(u) du

Where:

• g(x) is the new variable, and u = g(x)
• g'(x) is the derivative of g(x)

Example 1:

Evaluate the integral ∫ (2x + 3)^4 dx

Solution:

Let u = 2x + 3, then du = 2 dx

Substituting, we get:

∫ (2x + 3)^4 dx = 1/2 ∫ u^4 du

∫ (2x + 3)^4 dx = 1/2 (u^5/5) + C

∫ (2x + 3)^4 dx = 1/2 ((2x + 3)^5/5) + C

Example 2:

Evaluate the integral ∫ (1 + x^2)^(-1/2) dx

Solution:

Let u = 1 + x^2, then du = 2x dx

Substituting, we get:

∫ (1 + x^2)^(-1/2) dx = 1/2 ∫ u^(-1/2) du

∫ (1 + x^2)^(-1/2) dx = 1/2 (2u^(1/2)) + C

∫ (1 + x^2)^(-1/2) dx = √(1 + x^2) + C

Also Read: Profit and Loss Formula Questions

## Integration Formula by Partial Fractions

The Integration by Partial Fractions formula evaluates integrals where the integrand is a rational function, that is, a fraction with a polynomial in the numerator and a polynomial in the denominator.

The general Steps are:

1. Factorize the denominator of the rational function.
2. Express the rational function as a sum of partial fractions.
3. Integrate each partial fraction separately.

Example 1:

Evaluate the integral ∫ (x^2 + 3x + 2) / (x^2 – 1) dx

Solution:

Step 1: Factorize the denominator.

x^2 – 1 = (x – 1)(x + 1)

Step 2: Express the rational function as a sum of partial fractions.

(x^2 + 3x + 2) / (x^2 – 1) = A / (x – 1) + B / (x + 1)

Step 3: Integrate each partial fraction separately.

∫ (x^2 + 3x + 2) / (x^2 – 1) dx = ∫ A / (x – 1) dx + ∫ B / (x + 1) dx

∫ (x^2 + 3x + 2) / (x^2 – 1) dx = A ln|x – 1| + B ln|x + 1| + C

Example 2:

Evaluate the integral ∫ 1 / (x^3 – x) dx

Solution:

Step 1: Factorize the denominator.

x^3 – x = x(x^2 – 1) = x(x – 1)(x + 1)

Step 2: Express the rational function as a sum of partial fractions.

1 / (x^3 – x) = A / x + B / (x – 1) + C / (x + 1)

Step 3: Integrate each partial fraction separately.

∫ 1 / (x^3 – x) dx = ∫ A / x dx + ∫ B / (x – 1) dx + ∫ C / (x + 1) dx

∫ 1 / (x^3 – x) dx = A ln|x| + B ln|x – 1| – C ln|x + 1| + D

## Definite Integration Formula

The Definite Integration Formula is used to evaluate the integral of a function over a specific interval [a, b].

The Formula is:

∫_a^b f(x) dx = F(b) – F(a)

Where:

• F(x) is the antiderivative or indefinite integral of f(x)

Example:

Evaluate the integral ∫_0^2 (x^2 + 3x + 1) dx

Solution:

Step 1: Find the antiderivative of the integrand.

F(x) = ∫ (x^2 + 3x + 1) dx = x^3/3 + 3x^2/2 + x + C

Step 2: Evaluate the Definite Integral using the formula.

∫_0^2 (x^2 + 3x + 1) dx = F(2) – F(0)

∫_0^2 (x^2 + 3x + 1) dx = (2^3/3 + 3(2)^2/2 + 2 + C) – (0^3/3 + 3(0)^2/2 + 0 + C)

∫_0^2 (x^2 + 3x + 1) dx = 8/3 + 6 + 2 – 0 = 16/3

## Indefinite Integration Formula

The Indefinite Integration Formula is used to find the antiderivative or primitive function of a given function.

The Formula is:

∫ f(x) dx = F(x) + C

Where:

• F(x) is the antiderivative or primitive function of f(x)
• C is the constant of integration

Example:

Evaluate the integral ∫ (2x + 3) / (x^2 + 1) dx

Solution:

Step 1: Identify the integrand.

f(x) = (2x + 3) / (x^2 + 1)

Step 2: Find the antiderivative of the integrand.

∫ (2x + 3) / (x^2 + 1) dx = ∫ 2 / (x^2 + 1) dx + ∫ 3 / (x^2 + 1) dx

Using the Integration by Substitution Formula:

∫ 2 / (x^2 + 1) dx = 2 tan^-1(x) + C

∫ 3 / (x^2 + 1) dx = 3 tan^-1(x) + C

Combining the results:

∫ (2x + 3) / (x^2 + 1) dx = 2 tan^-1(x) + 3 tan^-1(x) + C

∫ (2x + 3) / (x^2 + 1) dx = 5 tan^-1(x) + C