Partial Fractions: Types, Methods and Solved Examples

Introduction

A partial fraction is a simpler rational fraction obtained by splitting a complicated rational expression into the sum of two or more simpler fractions.

The process of expressing a rational fraction as a sum of simpler fractions is called partial fraction decomposition.

For example:

5 / [x(x + 2)] = 5/(2x) − 5/[2(x + 2)]

Instead of working with one complicated fraction, we can work with two simpler fractions.

Partial fractions are useful in integration, differential equations and certain pharmaceutical calculations. Before applying this method, the given rational fraction should be proper. If it is improper, perform polynomial division first.

1. Basic Concept of Partial Fractions

Consider the following expression:

1 / [x(x + 1)]

The denominator contains two different linear factors: x and x + 1.

We can express the fraction as:

1 / [x(x + 1)] = A/x + B/(x + 1)

Here, A and B are unknown constants that we need to determine.

Multiplying both sides by x(x + 1) gives:

1 = A(x + 1) + Bx

Expanding:

1 = Ax + A + Bx

Rearranging:

1 = (A + B)x + A

Comparing the constant terms and coefficients of x:

A = 1

A + B = 0

Since A = 1, we get B = −1.

Therefore:

1 / [x(x + 1)] = 1/x − 1/(x + 1)

This is the basic principle of partial fraction decomposition.

2. Types of Partial Fractions

The form of the partial fractions depends on how the denominator can be factorized.

A. Distinct Linear Factors

When the denominator contains different linear factors, each factor receives a separate constant numerator.

For example, if the denominator is (x + 1)(x + 2), write:

P(x) / [(x + 1)(x + 2)] = A/(x + 1) + B/(x + 2)

B. Repeated Linear Factors

When a linear factor is repeated, include a separate fraction for every power of that factor.

For example:

P(x) / (x + 1)² = A/(x + 1) + B/(x + 1)²

For a factor repeated three times, include all three powers:

A/(x + 1) + B/(x + 1)² + C/(x + 1)³

C. Irreducible Quadratic Factors

An irreducible quadratic factor is a quadratic polynomial that cannot be factorized into real linear factors.

For a denominator containing x² + 1, the corresponding numerator must be linear:

(Ax + B) / (x² + 1)

The numerator contains both Ax and B, because a constant numerator alone is not general enough for every decomposition.

3. Solved Examples

Example 1 — Basic Level: Distinct Linear Factors

Question: Resolve the following into partial fractions:

5 / [x(x + 2)]

Step 1: Assume the decomposition.

5 / [x(x + 2)] = A/x + B/(x + 2)

Step 2: Multiply by the denominator.

5 = A(x + 2) + Bx

Expanding:

5 = Ax + 2A + Bx

5 = (A + B)x + 2A

Step 3: Compare coefficients.

Constant terms:

2A = 5

A = 5/2

Coefficient of x:

A + B = 0

5/2 + B = 0

B = −5/2

Final answer:

5 / [x(x + 2)] = 5/(2x) − 5/[2(x + 2)]

Example 2 — Medium Level: Repeated Linear Factor

Question: Resolve into partial fractions:

(3x + 5) / (x + 1)²

Step 1: Write the correct form.

Because x + 1 is repeated twice:

(3x + 5) / (x + 1)² = A/(x + 1) + B/(x + 1)²

Step 2: Multiply by (x + 1)².

3x + 5 = A(x + 1) + B

Expanding:

3x + 5 = Ax + A + B

Step 3: Compare coefficients.

Coefficient of x:

A = 3

Constant terms:

A + B = 5

Substituting A = 3:

3 + B = 5

B = 2

Final answer:

(3x + 5) / (x + 1)² = 3/(x + 1) + 2/(x + 1)²

Example 3 — Challenging Level: Irreducible Quadratic Factor

Question: Resolve into partial fractions:

(2x² + 3x + 4) / [(x + 1)(x² + 1)]

Step 1: Write the decomposition.

The denominator contains a linear factor and an irreducible quadratic factor.

Therefore:

(2x² + 3x + 4) / [(x + 1)(x² + 1)] = A/(x + 1) + (Bx + C)/(x² + 1)

Step 2: Multiply by the denominator.

2x² + 3x + 4 = A(x² + 1) + (Bx + C)(x + 1)

Expanding:

2x² + 3x + 4 = Ax² + A + Bx² + Bx + Cx + C

Collecting terms:

2x² + 3x + 4 = (A + B)x² + (B + C)x + (A + C)

Step 3: Compare coefficients.

A + B = 2

B + C = 3

A + C = 4

From the first equation:

B = 2 − A

From the second:

C = 3 − B = 1 + A

Substitute into the third equation:

A + (1 + A) = 4

2A = 3

A = 3/2

Therefore:

B = 2 − 3/2 = 1/2

C = 1 + 3/2 = 5/2

Final answer:

(2x² + 3x + 4) / [(x + 1)(x² + 1)] = 3/[2(x + 1)] + (x + 5)/[2(x² + 1)]

4. Important Rules to Remember

• If the fraction is improper, perform polynomial division before decomposition.

• For each distinct linear factor, use a constant numerator.

• For a repeated linear factor, include every power up to the highest repetition.

• For an irreducible quadratic factor, use a linear numerator of the form Ax + B.

• After finding the constants, combine the fractions to verify that the original expression is recovered.

5. Practice Questions

Try resolving these rational fractions into partial fractions.

Basic:

3 / [x(x + 1)]

Medium:

(2x + 7) / (x + 2)²

Challenging:

(x² + 2x + 3) / [(x + 1)(x² + 1)]

Answers for Checking

• Basic: 3/x − 3/(x + 1)

• Medium: 2/(x + 2) + 3/(x + 2)²

• Challenging: 1/(x + 1) + 1/(x² + 1)

Key Takeaways

Partial fraction decomposition breaks a rational expression into simpler fractions. The denominator's factorization determines the correct form of the decomposition, and unknown constants are found by comparing coefficients or substituting convenient values of the variable.

This method is an important foundation for the next steps in Remedial Mathematics, particularly integration and differential equations.