Unit 6.12 – Using Linear Partial Fractions

AP® Calculus BC ONLY | Decomposing Rational Functions for Integration

Why This Matters: Partial fraction decomposition is a powerful algebraic technique that breaks complex rational functions into simpler fractions that are easier to integrate! When you have a rational function (polynomial over polynomial) that can't be integrated directly, partial fractions decomposes it into a sum of simpler fractions. This BC-only topic is essential for integrating rational functions and appears frequently on the AP® exam. Master this technique—it's one of the most elegant problem-solving tools in calculus!

🎯 What Are Partial Fractions?

THE CONCEPT

Partial fraction decomposition reverses the process of adding fractions!

Example of the Reverse Process:

Adding fractions (forward):

\[ \frac{2}{x-1} + \frac{3}{x+2} = \frac{2(x+2) + 3(x-1)}{(x-1)(x+2)} = \frac{5x+1}{x^2+x-2} \]

Partial fractions (reverse):

\[ \frac{5x+1}{x^2+x-2} = \frac{2}{x-1} + \frac{3}{x+2} \]

When to Use Partial Fractions:

Integrating rational functions: Where direct methods don't work
Denominator factors nicely: Into linear factors (BC scope)
Proper fractions: If improper, use long division first!

⚠️ Important: For AP® BC, you only need LINEAR factors (factors of the form \(ax + b\)). Irreducible quadratic factors are beyond BC scope!

📋 Types of Linear Partial Fraction Decomposition

Linear Partial Fraction Forms (BC Only)

ptionion>Decomposition Patterns

Case	Denominator Form	Partial Fraction Setup
Case 1	Distinct linear factors \((x-a)(x-b)\)	\(\frac{A}{x-a} + \frac{B}{x-b}\)
Case 2	Repeated linear factor \((x-a)^2\)	\(\frac{A}{x-a} + \frac{B}{(x-a)^2}\)
Case 3	Triple repeated factor \((x-a)^3\)	\(\frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{(x-a)^3}\)

🔧 The Partial Fractions Method

Step-by-Step Procedure:

Check if proper: Is deg(numerator) < deg(denominator)?
- If NO: Use polynomial long division first!
- If YES: Proceed to next step
Factor denominator: Completely into linear factors
Set up partial fraction form: Based on factor type
- Distinct factors: one term per factor
- Repeated factors: one term for each power
Clear denominators: Multiply both sides by LCD
Solve for constants: Use substitution or system of equations
- Method A: Strategic substitution (easier!)
- Method B: Expand and equate coefficients
Write decomposition: Substitute values back
Integrate: Each simple fraction separately

📖 Comprehensive Worked Examples

Example 1: Distinct Linear Factors

Problem: Find \(\int \frac{5x+1}{x^2+x-2} \, dx\)

Solution:

Step 1: Check if proper

Numerator degree: 1, Denominator degree: 2

Since 1 < 2, it's proper ✓

Step 2: Factor denominator

\[ x^2 + x - 2 = (x-1)(x+2) \]

Step 3: Set up partial fractions

\[ \frac{5x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} \]

Step 4: Clear denominators

Multiply both sides by \((x-1)(x+2)\):

\[ 5x + 1 = A(x+2) + B(x-1) \]

Step 5: Solve for A and B (Strategic Substitution)

Let \(x = 1\):

\[ 5(1) + 1 = A(1+2) + B(0) \]

\[ 6 = 3A \quad \Rightarrow \quad A = 2 \]

Let \(x = -2\):

\[ 5(-2) + 1 = A(0) + B(-2-1) \]

\[ -9 = -3B \quad \Rightarrow \quad B = 3 \]

Step 6: Write decomposition

\[ \frac{5x+1}{x^2+x-2} = \frac{2}{x-1} + \frac{3}{x+2} \]

Step 7: Integrate

\[ \int \frac{5x+1}{x^2+x-2} \, dx = \int \frac{2}{x-1}\,dx + \int \frac{3}{x+2}\,dx \]

\[ = 2\ln|x-1| + 3\ln|x+2| + C \]

Answer: \(2\ln|x-1| + 3\ln|x+2| + C\)

Example 2: Three Distinct Linear Factors

Problem: Find \(\int \frac{2x^2 - x + 4}{x(x-1)(x+2)} \, dx\)

Solution:

Step 1-2: Proper fraction, already factored ✓

Step 3: Set up partial fractions

\[ \frac{2x^2 - x + 4}{x(x-1)(x+2)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+2} \]

Step 4: Clear denominators

\[ 2x^2 - x + 4 = A(x-1)(x+2) + Bx(x+2) + Cx(x-1) \]

Step 5: Strategic substitution

Let \(x = 0\):

\[ 4 = A(-1)(2) \quad \Rightarrow \quad A = -2 \]

Let \(x = 1\):

\[ 2 - 1 + 4 = B(1)(3) \quad \Rightarrow \quad 5 = 3B \quad \Rightarrow \quad B = \frac{5}{3} \]

Let \(x = -2\):

\[ 8 + 2 + 4 = C(-2)(-3) \quad \Rightarrow \quad 14 = 6C \quad \Rightarrow \quad C = \frac{7}{3} \]

Step 6-7: Decomposition and integration

\[ = -\frac{2}{x} + \frac{5/3}{x-1} + \frac{7/3}{x+2} \]

\[ \int = -2\ln|x| + \frac{5}{3}\ln|x-1| + \frac{7}{3}\ln|x+2| + C \]

Answer: \(-2\ln|x| + \frac{5}{3}\ln|x-1| + \frac{7}{3}\ln|x+2| + C\)

Example 3: Repeated Linear Factor

Problem: Find \(\int \frac{3x+5}{(x-1)^2} \, dx\)

Solution:

Step 1-2: Proper, already factored ✓

Step 3: Set up (repeated factor!)

\[ \frac{3x+5}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2} \]

Note: Need terms for both powers!

Step 4: Clear denominators

Multiply by \((x-1)^2\):

\[ 3x + 5 = A(x-1) + B \]

Step 5: Solve for A and B

Let \(x = 1\):

\[ 3(1) + 5 = A(0) + B \quad \Rightarrow \quad B = 8 \]

Let \(x = 0\): (or expand and compare)

\[ 5 = A(-1) + 8 \quad \Rightarrow \quad A = 3 \]

Step 6-7: Integrate

\[ \int \frac{3x+5}{(x-1)^2} \, dx = \int \frac{3}{x-1}\,dx + \int \frac{8}{(x-1)^2}\,dx \]

\[ = 3\ln|x-1| + 8 \cdot \frac{(x-1)^{-1}}{-1} + C \]

\[ = 3\ln|x-1| - \frac{8}{x-1} + C \]

Answer: \(3\ln|x-1| - \frac{8}{x-1} + C\)

Example 4: Repeated Factor with Distinct Factor

Problem: Find \(\int \frac{x^2 + 1}{x(x+1)^2} \, dx\)

Solution:

Step 3: Set up partial fractions

\[ \frac{x^2 + 1}{x(x+1)^2} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} \]

Step 4: Clear denominators

\[ x^2 + 1 = A(x+1)^2 + Bx(x+1) + Cx \]

Step 5: Strategic substitution

Let \(x = 0\):

\[ 1 = A(1) \quad \Rightarrow \quad A = 1 \]

Let \(x = -1\):

\[ 1 + 1 = C(-1) \quad \Rightarrow \quad C = -2 \]

Let \(x = 1\): (to find B)

\[ 2 = A(4) + B(2) + C(1) = 4 + 2B - 2 \]

\[ 2 = 2 + 2B \quad \Rightarrow \quad B = 0 \]

Step 6-7: Integrate

\[ \int \frac{x^2+1}{x(x+1)^2}\,dx = \int \frac{1}{x}\,dx - 2\int \frac{1}{(x+1)^2}\,dx \]

\[ = \ln|x| + \frac{2}{x+1} + C \]

Answer: \(\ln|x| + \frac{2}{x+1} + C\)

Example 5: With Long Division First

Problem: Find \(\int \frac{x^3 + 2x + 1}{x^2 - 1} \, dx\)

Solution:

Step 1: Check if proper

Numerator degree 3 > denominator degree 2 → IMPROPER!

Must use long division first!

Long division:

\[ \frac{x^3 + 2x + 1}{x^2 - 1} = x + \frac{3x + 1}{x^2 - 1} \]

Now apply partial fractions to remainder:

Factor: \(x^2 - 1 = (x-1)(x+1)\)

\[ \frac{3x+1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} \]

Solving: \(A = 2, B = 1\)

Integrate:

\[ \int \left(x + \frac{2}{x-1} + \frac{1}{x+1}\right)dx \]

\[ = \frac{x^2}{2} + 2\ln|x-1| + \ln|x+1| + C \]

Answer: \(\frac{x^2}{2} + 2\ln|x-1| + \ln|x+1| + C\)

💡 Essential Tips & Strategies

✅ Setting Up Success:

Always check proper vs improper: If deg(num) ≥ deg(den), use long division first!
Factor completely: Make sure denominator is fully factored
Repeated factors: Include terms for ALL powers up to the multiplicity
One constant per term: Each partial fraction gets its own constant (A, B, C, ...)

🔥 Solving for Constants:

Strategic substitution is faster: Plug in values that make factors zero
Choose x values wisely: Use zeros of linear factors to eliminate variables
If stuck: Expand and equate coefficients (more work, but always works)
Check your answer: Add fractions back to verify

Common Decomposition Patterns:

Quick Reference Guide
Denominator	Partial Fraction Form
\((x-a)(x-b)\)	\(\frac{A}{x-a} + \frac{B}{x-b}\)
\((x-a)^2\)	\(\frac{A}{x-a} + \frac{B}{(x-a)^2}\)
\((x-a)^3\)	\(\frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{(x-a)^3}\)
\(x(x-a)(x-b)\)	\(\frac{A}{x} + \frac{B}{x-a} + \frac{C}{x-b}\)
\(x(x-a)^2\)	\(\frac{A}{x} + \frac{B}{x-a} + \frac{C}{(x-a)^2}\)

❌ Common Mistakes to Avoid

Mistake 1: Not checking if fraction is proper before starting
Mistake 2: For repeated factors, forgetting to include ALL powers
Mistake 3: Not factoring denominator completely
Mistake 4: Algebra errors when clearing denominators
Mistake 5: Sign errors when integrating \(\frac{1}{(x-a)^n}\)
Mistake 6: Forgetting absolute value in \(\ln|x-a|\)
Mistake 7: Not using strategic substitution (makes work much harder)
Mistake 8: Forgetting +C at the end
Mistake 9: Incorrect integration of \((x-a)^{-n}\) terms
Mistake 10: Not verifying decomposition by adding fractions back

📐 Integration Formulas You'll Need

After Partial Fraction Decomposition:

Function Form	Integral
\(\int \frac{A}{x-a}\,dx\)	\(A\ln\|x-a\| + C\)
\(\int \frac{B}{(x-a)^2}\,dx\)	\(-\frac{B}{x-a} + C\)
\(\int \frac{C}{(x-a)^3}\,dx\)	\(-\frac{C}{2(x-a)^2} + C\)
\(\int \frac{D}{(x-a)^n}\,dx\)	\(\frac{D}{(1-n)(x-a)^{n-1}} + C\) (for \(n \neq 1\))

📝 Practice Problems

Set A: Basic Partial Fractions

\(\int \frac{7}{x^2-4} \, dx\)
\(\int \frac{x+5}{x^2+3x+2} \, dx\)
\(\int \frac{1}{x(x-3)} \, dx\)

Hints:

Factor: \(x^2-4 = (x-2)(x+2)\)
Factor: \(x^2+3x+2 = (x+1)(x+2)\)
Already factored

Answers:

\(\frac{7}{4}\ln\left|\frac{x-2}{x+2}\right| + C\)
\(3\ln|x+1| - 2\ln|x+2| + C\)
\(-\frac{1}{3}\ln|x| + \frac{1}{3}\ln|x-3| + C\)

Set B: Repeated Factors

\(\int \frac{2x-1}{(x+1)^2} \, dx\)
\(\int \frac{x+1}{x^2(x-1)} \, dx\)

Answers:

\(2\ln|x+1| + \frac{3}{x+1} + C\)
\(-\frac{1}{x} + 2\ln|x| - \ln|x-1| + C\)

✏️ AP® Exam Success Tips

What AP® BC Graders Look For:

Show partial fraction setup: Write the decomposition form clearly
Show one equation: The equation after clearing denominators
Show solving process: At least one substitution to find constants
Write final decomposition: With numerical values
Integrate each term: Show integration of each simple fraction
Include +C: Don't forget constant of integration
If improper: Show long division step

⚡ Ultimate Quick Reference

PARTIAL FRACTIONS CHECKLIST

The 7-Step Process:

✓ Check if proper (deg num < deg den)
✓ Factor denominator completely
✓ Set up partial fraction form
✓ Clear denominators (multiply by LCD)
✓ Solve for constants (strategic substitution!)
✓ Write decomposition with values
✓ Integrate each term, add +C

Key Patterns:

Distinct: \(\frac{A}{x-a} + \frac{B}{x-b}\)
Repeated: \(\frac{A}{x-a} + \frac{B}{(x-a)^2}\)
Triple: Add \(\frac{C}{(x-a)^3}\)

Master Partial Fractions! This BC-only technique decomposes rational functions into simpler fractions for integration. Step 1: Check if proper (numerator degree < denominator degree)—if not, use long division first. Step 2: Factor denominator into linear factors. Step 3: Set up decomposition: distinct factors \((x-a)\) get \(\frac{A}{x-a}\); repeated factors \((x-a)^n\) need terms for all powers: \(\frac{A}{x-a} + \frac{B}{(x-a)^2} + \cdots + \frac{N}{(x-a)^n}\). Step 4: Clear denominators by multiplying by LCD. Step 5: Use strategic substitution—plug in values that make factors zero to easily solve for constants. Step 6: Write decomposition with values. Step 7: Integrate: \(\int \frac{A}{x-a}\,dx = A\ln|x-a| + C\) and \(\int \frac{B}{(x-a)^n}\,dx = \frac{B}{(1-n)(x-a)^{n-1}} + C\). For BC, only linear factors are tested—no irreducible quadratics! Always verify by adding fractions back. This appears frequently on BC exams—practice until automatic! 🎯✨