📐 Concept diagram

05-07 - Partial Fractions Integration

Phase: 5 | Subject: 05-07 Prerequisites: 05-04-integration-by-substitution.md, 03-04-rational-functions.md Next subject: 05-08-improper-integrals.md

Learning Objectives

By the end of this subject, you will be able to:

1. Decompose proper rational functions into partial fractions
2. Handle linear factors (distinct and repeated)
3. Handle irreducible quadratic factors
4. Integrate rational functions using partial fractions

Core Content

Why Partial Fractions?

Many rational functions don't have elementary antiderivatives in their original form, but AFTER decomposition, each term does.

Example: ∫1/(x² - 1)dx = ∫[1/(2(x-1)) - 1/(2(x+1))]dx = (1/2)ln|x - 1| - (1/2)ln|x + 1| + C = (1/2)ln|(x - 1)/(x + 1)| + C

Decomposition Rules

Linear Factors (Distinct)

For each factor (x - a): term is A/(x - a)

Example: 1/((x - 1)(x + 2)) = A/(x - 1) + B/(x + 2)

Repeated Linear Factors

For (x - a)ⁿ: terms are A₁/(x - a) + A₂/(x - a)² + ... + Aₙ/(x - a)ⁿ

Example: 1/((x - 1)²(x + 2)) = A/(x - 1) + B/(x - 1)² + C/(x + 2)

Irreducible Quadratic Factors

For (ax² + bx + c) where b² - 4ac < 0: term is (Ax + B)/(ax² + bx + c)

Example: 1/((x - 1)(x² + 1)) = A/(x - 1) + (Bx + C)/(x² + 1)

Finding Coefficients

Multiply through by the denominator, then: - Plug in roots of the denominator (for linear factors) - Equate coefficients (for remaining unknowns)

Key Terms

• 05 07 Partial Fractions Integration
• Correct: B)
• Correct: C)
• Decomposition Rules
• Example 1: Distinct linear factors
• Example 2: Repeated factors
• Example 3: Quadratic factor
• Finding Coefficients
• Irreducible Quadratic Factors
• Linear Factors (Distinct)
• Repeated Linear Factors
• Why Partial Fractions?

Worked Examples

Example 1: Distinct linear factors

Decompose 1/((x - 1)(x + 3))

1 = A(x + 3) + B(x - 1) x = 1: 1 = 4A, A = 1/4 x = -3: 1 = -4B, B = -1/4

Result: 1/(4(x - 1)) - 1/(4(x + 3))

Integral: (1/4)ln|x - 1| - (1/4)ln|x + 3| + C

Example 2: Repeated factors

Decompose 1/((x - 2)²(x + 1))

1 = A(x - 2)(x + 1) + B(x + 1) + C(x - 2)² x = 2: 1 = 3B, B = 1/3 x = -1: 1 = 9C, C = 1/9

Expand and compare coefficients for A: 1 = A(x² - x - 2) + (1/3)(x + 1) + (1/9)(x² - 4x + 4) Collect x²: 0 = A + 1/9, so A = -1/9

Example 3: Quadratic factor

Decompose 1/((x + 1)(x² + 1))

1 = A(x² + 1) + (Bx + C)(x + 1) x = -1: 1 = 2A, A = 1/2 Compare x²: 0 = A + B, so B = -1/2 Compare constants: 1 = A + C, so C = 1/2

Result: 1/(2(x + 1)) + (-x/2 + 1/2)/(x² + 1)

Practice Problems

Problem 1: Decompose 1/(x(x + 2))

Problem 2: ∫1/(x² - 9)dx

Problem 3: ∫1/((x+1)²)dx

Problem 4: ∫(2x+1)/(x²+x-6)dx

Problem 5: ∫(x³+1)/(x²−1)dx

Summary

Key takeaways:

• Decompose proper rational functions: degree numerator < degree denominator
• Linear factors → A/(x - a)
• Repeated factors → sum of decreasing powers
• Quadratic factors → (Ax + B)/(quadratic)
• Integrate each term separately

Pitfalls

• Forgetting to check the degree of the numerator. Partial fraction decomposition requires deg(numerator) < deg(denominator). If deg(num) ≥ deg(den), polynomial long division must be performed first. Skipping this step gives an incorrect decomposition.
• Missing terms for repeated factors. For a factor (x − a)ⁿ, you need n terms: A₁/(x−a) + A₂/(x−a)² + ... + Aₙ/(x−a)ⁿ. Using only A/(x−a)ⁿ is insufficient — the lower-power terms are required to match all degrees of freedom.
• Using a constant numerator for irreducible quadratics. For an irreducible quadratic factor ax² + bx + c, the numerator must be linear: (Ax + B)/(ax² + bx + c). A constant numerator A alone cannot represent all possible rational functions.
• Arithmetic errors when solving for coefficients. After setting up e.g. 1 = A(x+3) + B(x−1), arithmetic mistakes when plugging in root values (x = 1, x = −3) are common. Always verify by combining the decomposition back into a single fraction.
• Forgetting how to integrate terms from quadratic factors. Terms like (Ax + B)/(x² + a²) split into two: Ax/(x² + a²) integrates to (A/2)ln|x² + a²|, and B/(x² + a²) integrates to (B/a)arctan(x/a). Missing this split leaves the integral unsolved.

Quiz

Q1: The partial fraction decomposition of 1/((x-1)(x+2)) is:

A) 1/(x-1) + 1/(x+2) B) 1/(3(x-1)) - 1/(3(x+2)) C) 1/(x-1) - 1/(x+2) D) 1/(3(x-1)) + 1/(3(x+2))

Q2: For 1/((x-1)²), the decomposition is:

A) 1/(x-1) B) 1/(x-1)² C) A/(x-1) + B/(x-1)² D) A/(x-1)² + B/(x-1)

Q3: ∫1/(x² + 1)dx equals:

A) ln|x² + 1| + C B) arctan(x) + C C) (1/2)ln|x² + 1| + C D) 1/x + C

Q4: ∫1/(x² - 4)dx equals:

A) ln|x² - 4| + C B) (1/4)ln|(x-2)/(x+2)| + C C) (1/2)ln|(x-2)/(x+2)| + C D) arctan(x/2) + C

Q5: When the denominator has an irreducible quadratic factor x² + 1, the partial fraction term is:

A) A/(x² + 1) B) A/(x² + 1)² C) (Ax + B)/(x² + 1) D) A·x + B

Next Steps

Next up: 05-08-improper-integrals.md