When should I use partial fraction decomposition?

You should use partial fraction decomposition when you need to integrate rational functions where the degree of the numerator is less than the degree of the denominator.

How do I decompose a rational function into partial fractions?

To decompose a rational function, first factor the denominator into irreducible polynomials, then express the function as a sum of simpler fractions and solve for the constants.

What types of denominators require partial fraction decomposition?

Denominators that are products of linear factors, repeated linear factors, or irreducible quadratic factors typically require partial fraction decomposition.

What if the numerator degree is greater than or equal to the denominator?

If the numerator degree is greater than or equal to the denominator degree, perform polynomial long division first, then apply partial fractions to the resulting proper fraction.

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Integration by Partial Fractions

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Integration by Partial Fractions

Q: What is integration by partial fractions?

Integration by partial fractions is a technique used to break down complex rational functions into simpler fractions, which can then be easily integrated.

Integration by partial fractions is a powerful technique used to simplify the integration of rational functions. A rational function is the quotient of 2 polynomials, with the degree of the numerator being less than or equal to the degree of the denominator. When faced with such an expression, breaking it down into simpler fractions makes the integration process much easier.

1.0What is Integration by Partial Fractions?

The main idea behind Integration by Partial Fractions is to decompose a complex rational function into simpler fractions that are easier to integrate. The method is most commonly used when the denominator can be factored into simpler polynomials. By expressing the function as a sum of simpler fractions, each with a simpler denominator, we can integrate each fraction separately.

For example, consider a rational function like:

$\frac{P ( x )}{Q ( x )}$

where P(x) and Q(x) are polynomials, and Q(x) can be factored into simpler terms. We then express the rational function as a sum of partial fractions.

2.0The Integration by Partial Fractions Formula

Rational function is defined as the ratio of two polynomials in the form $\frac{P ( x )}{Q ( x )}$ , where P(x) and Q(x) are polynomials in x and $Q (x) \neq = 0$ . If the degree of P(x) is less than the degree of Q(x), then the rational function is called proper, otherwise, it is called improper. The improper rational function can be reduced to the proper rational functions by a long division process. Thus, if $\frac{P ( x )}{Q ( x )}$ is improper, then $\frac{P ( x )}{Q ( x )} = T (x) + \frac{P _{1} ( x )}{Q ( x )}$ , where T(x) is a polynomial in x and $\frac{P _{1} ( x )}{Q ( x )}$ is a proper rational function. It is always possible to write the integrand as a sum of simpler rational functions by a method called partial fraction decomposition. After this, the integration can be carried out easily using the already known methods.

S. No.	Form of the rational function	Form of the partial fraction
1.	$\frac{p x ^{2} + q x + r}{( x - a ) ( x - b ) ( x - c )}$	$\frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c}$
2.	$\frac{p x ^{2} + q x + r}{( x - a ) ^{2} ( x - b )}$	$\frac{A}{x - a} + \frac{B}{( x - a ) ^{2}} + \frac{C}{x - b}$
3.	$\frac{p x ^{2} + q x + r}{( x - a ) ( x ^{2} + b x + c )}$	$\frac{A}{x - a} + \frac{B x + C}{x ^{2} + b x + c}$

where $x^{2} + b x + c$ cannot be factorised further

3.0Steps For Integration By Partial Fractions

To integrate a rational function using partial fractions, follow these steps:

Factor the Denominator: Begin by factoring the denominator of the rational function completely. If it is already factored, skip this step.
Set Up the Partial Fraction Decomposition: Based on the factors of the denominator, set up the appropriate partial fractions.
Solve for the Constants: Once you have the decomposition, multiply both sides of the equation by the denominator and solve for the constants (like A, B, etc.).
Integrate Each Fraction: After determining the constants, you can now integrate each fraction separately using basic integration rules.
Combine the Results: Finally, combine the results of the individual integrals to obtain the solution to the original integral.

4.0Solved Example of Integration by Partial Fractions

Example 1: Evaluate: $\int \frac{x}{( x - 2 ) ( x + 5 )} d x$

Solution:

$\frac{x}{( x - 2 ) ( x + 5 )} = \frac{A}{x - 2} + \frac{B}{x + 5}$

or $x = A (x + 5) + B (x -2)$ .

by comparing the coefficients, we get

A = 2/7 and B = 5/7 so that

$\int \frac{x}{( x - 2 ) ( x + 5 )} d x$ $= \frac{2}{7} \int \frac{d x}{x - 2} + \frac{5}{7} \int d x x + 5$

$= \frac{2}{7} l n ∣ (x - 2) ∣ + \frac{5}{7} l n ∣ (x + 5) ∣ + C$ = Ans.

Example 2: Evaluate $\int \frac{x ^{4}}{( x + 2 ) ( x ^{2} + 1 )} d x$

Solution:

$\frac{x ^{4}}{( x + 2 ) ( x ^{2} + 1 )}$ $= (x - 2) + \frac{3 x ^{2} + 4}{( x + 2 ) ( x ^{2} + 1 )}$

Now, $\frac{3 x ^{2} + 4}{( x + 2 ) ( x ^{2} + 1 )}$ $= \frac{16}{5 ( x + 2 )} + \frac{- \frac{1}{5} x + \frac{2}{5}}{x ^{2} + 1}$

So, $\frac{x ^{4}}{( x + 2 ) ( x ^{2} + 1 )}$ $= x - 2 + \frac{16}{5 ( x + 2 )} + \frac{- \frac{1}{5} x + \frac{2}{5}}{x ^{2} + 1}$

$\int [(x - 2) + \frac{16}{5 ( x + 2 )} + \frac{- \frac{1}{5} x + \frac{2}{5}}{x ^{2} + 1}]$

Now,

$\frac{x ^{2}}{2} - 2 x + \frac{2}{5} t a n^{- 1} x + \frac{16}{5} l n ∣ x + 2∣ - \frac{1}{10} l n (x^{2} + 1) + C$ = Ans.

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Integration by Partial Fractions

Integration by partial fractions is a powerful technique used to simplify the integration of rational functions. A rational function is the quotient of 2 polynomials, with the degree of the numerator being less than or equal to the degree of the denominator. When faced with such an expression, breaking it down into simpler fractions makes the integration process much easier.

1.0What is Integration by Partial Fractions?

The main idea behind Integration by Partial Fractions is to decompose a complex rational function into simpler fractions that are easier to integrate. The method is most commonly used when the denominator can be factored into simpler polynomials. By expressing the function as a sum of simpler fractions, each with a simpler denominator, we can integrate each fraction separately.

For example, consider a rational function like:

$\frac{P ( x )}{Q ( x )}$

where P(x) and Q(x) are polynomials, and Q(x) can be factored into simpler terms. We then express the rational function as a sum of partial fractions.

2.0The Integration by Partial Fractions Formula

Rational function is defined as the ratio of two polynomials in the form $\frac{P ( x )}{Q ( x )}$ , where P(x) and Q(x) are polynomials in x and $Q (x) \neq = 0$ . If the degree of P(x) is less than the degree of Q(x), then the rational function is called proper, otherwise, it is called improper. The improper rational function can be reduced to the proper rational functions by a long division process. Thus, if $\frac{P ( x )}{Q ( x )}$ is improper, then $\frac{P ( x )}{Q ( x )} = T (x) + \frac{P _{1} ( x )}{Q ( x )}$ , where T(x) is a polynomial in x and $\frac{P _{1} ( x )}{Q ( x )}$ is a proper rational function. It is always possible to write the integrand as a sum of simpler rational functions by a method called partial fraction decomposition. After this, the integration can be carried out easily using the already known methods.

S. No.	Form of the rational function	Form of the partial fraction
1.	$\frac{p x ^{2} + q x + r}{( x - a ) ( x - b ) ( x - c )}$	$\frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c}$
2.	$\frac{p x ^{2} + q x + r}{( x - a ) ^{2} ( x - b )}$	$\frac{A}{x - a} + \frac{B}{( x - a ) ^{2}} + \frac{C}{x - b}$
3.	$\frac{p x ^{2} + q x + r}{( x - a ) ( x ^{2} + b x + c )}$	$\frac{A}{x - a} + \frac{B x + C}{x ^{2} + b x + c}$

where $x^{2} + b x + c$ cannot be factorised further

3.0Steps For Integration By Partial Fractions

To integrate a rational function using partial fractions, follow these steps:

Factor the Denominator: Begin by factoring the denominator of the rational function completely. If it is already factored, skip this step.
Set Up the Partial Fraction Decomposition: Based on the factors of the denominator, set up the appropriate partial fractions.
Solve for the Constants: Once you have the decomposition, multiply both sides of the equation by the denominator and solve for the constants (like A, B, etc.).
Integrate Each Fraction: After determining the constants, you can now integrate each fraction separately using basic integration rules.
Combine the Results: Finally, combine the results of the individual integrals to obtain the solution to the original integral.

4.0Solved Example of Integration by Partial Fractions

Example 1: Evaluate: $\int \frac{x}{( x - 2 ) ( x + 5 )} d x$

Solution:

$\frac{x}{( x - 2 ) ( x + 5 )} = \frac{A}{x - 2} + \frac{B}{x + 5}$

or $x = A (x + 5) + B (x -2)$ .

by comparing the coefficients, we get

A = 2/7 and B = 5/7 so that

$\int \frac{x}{( x - 2 ) ( x + 5 )} d x$ $= \frac{2}{7} \int \frac{d x}{x - 2} + \frac{5}{7} \int d x x + 5$

$= \frac{2}{7} l n ∣ (x - 2) ∣ + \frac{5}{7} l n ∣ (x + 5) ∣ + C$ = Ans.

Example 2: Evaluate $\int \frac{x ^{4}}{( x + 2 ) ( x ^{2} + 1 )} d x$

Solution:

$\frac{x ^{4}}{( x + 2 ) ( x ^{2} + 1 )}$ $= (x - 2) + \frac{3 x ^{2} + 4}{( x + 2 ) ( x ^{2} + 1 )}$

Now, $\frac{3 x ^{2} + 4}{( x + 2 ) ( x ^{2} + 1 )}$ $= \frac{16}{5 ( x + 2 )} + \frac{- \frac{1}{5} x + \frac{2}{5}}{x ^{2} + 1}$

So, $\frac{x ^{4}}{( x + 2 ) ( x ^{2} + 1 )}$ $= x - 2 + \frac{16}{5 ( x + 2 )} + \frac{- \frac{1}{5} x + \frac{2}{5}}{x ^{2} + 1}$

$\int [(x - 2) + \frac{16}{5 ( x + 2 )} + \frac{- \frac{1}{5} x + \frac{2}{5}}{x ^{2} + 1}]$

Now,

$\frac{x ^{2}}{2} - 2 x + \frac{2}{5} t a n^{- 1} x + \frac{16}{5} l n ∣ x + 2∣ - \frac{1}{10} l n (x^{2} + 1) + C$ = Ans.

Frequently Asked Questions

What is integration by partial fractions?

When should I use partial fraction decomposition?

How do I decompose a rational function into partial fractions?

What types of denominators require partial fraction decomposition?

What if the numerator degree is greater than or equal to the denominator?

Join ALLEN!

Integration by Partial Fractions

1.0What is Integration by Partial Fractions?

2.0The Integration by Partial Fractions Formula

3.0Steps For Integration By Partial Fractions

4.0Solved Example of Integration by Partial Fractions

Table of Contents

Frequently Asked Questions

What is integration by partial fractions?

When should I use partial fraction decomposition?

How do I decompose a rational function into partial fractions?

What types of denominators require partial fraction decomposition?

What if the numerator degree is greater than or equal to the denominator?

Join ALLEN!

Integration by Partial Fractions

1.0What is Integration by Partial Fractions?

2.0The Integration by Partial Fractions Formula

3.0Steps For Integration By Partial Fractions

4.0Solved Example of Integration by Partial Fractions

Table of Contents