📚 Partial Fractions Study Notes

Complete Guide for IB Mathematics AA HL

Theory, Examples & Interactive Practice

📖 Study Guide Navigation

📌 What are Partial Fractions?

🔢 Types of Partial Fractions

⚙️ Step-by-Step Method

📐 Linear Factors (Distinct)

🔁 Repeated Linear Factors

📊 Quadratic Factors

⚡ Improper Fractions

✍️ Interactive Practice Problems

📌 What are Partial Fractions?

Partial fractions are the reverse process of adding or subtracting algebraic fractions. Instead of combining fractions into a single fraction, we decompose a complex fraction into simpler parts.

Example: (3x + 5)/(2x² - 5x - 3) = 2/(x - 3) - 1/(2x + 1)

This technique is essential for integration, differential equations, and simplifying complex rational expressions in IB Mathematics AA HL.

🎯 Why Study Partial Fractions in IB?

Integration Technique: Essential for integrating rational functions that don't fit standard forms
Exam Weight: Regular appearance in IB Mathematics AA HL Paper 1 and Paper 2 (typically 4-8 marks)
Prerequisites: Requires strong factoring skills, algebraic manipulation, and equation solving
Applications: Used in solving differential equations, Laplace transforms, and complex analysis
Connection to Other Topics: Links to polynomial division, logarithmic integration, and series expansions

⚠️ Key Prerequisite: Before attempting partial fractions, ensure you're confident with factoring quadratic and cubic expressions, solving simultaneous equations, and algebraic manipulation. These skills are fundamental to success with this topic.

🔢 Types of Partial Fractions

The form of partial fractions depends on the factors in the denominator. Here are the main types you'll encounter in IB Mathematics:

Type 1: Distinct Linear Factors

P(x)/((ax + b)(cx + d)) = A/(ax + b) + B/(cx + d)

Where A and B are constants to be determined.

Type 2: Repeated Linear Factors

P(x)/(ax + b)² = A/(ax + b) + B/(ax + b)²

Each power requires a separate term in the decomposition.

Type 3: Quadratic Factor (Irreducible)

P(x)/((ax + b)(cx² + dx + e)) = A/(ax + b) + (Bx + C)/(cx² + dx + e)

Quadratic factors require linear numerators (Bx + C).

Type 4: General Repeated Factor

P(x)/(ax + b)³ = A/(ax + b) + B/(ax + b)² + C/(ax + b)³

For (ax + b)ⁿ, include terms up to power n.

💡 Recognition Tip: Always factorize the denominator completely first. The type of partial fraction decomposition is determined entirely by the factors in the denominator, not the numerator.

⚙️ Step-by-Step Method for Partial Fractions

Follow this systematic approach for any partial fraction problem:

STEP 1: Check if Proper Fraction

Ensure the degree of the numerator is less than the degree of the denominator. If not, perform polynomial long division first to make it proper.

Proper: (3x + 5)/(x² + 2x - 3) ✓ Improper: (x³ + 2)/(x² - 1) ✗

STEP 2: Factorize the Denominator

Completely factorize the denominator into linear and/or quadratic factors. Simplify if any common factors exist between numerator and denominator.

Example: 2x² - 5x - 3 = (2x + 1)(x - 3)

STEP 3: Set Up Partial Fractions

Write the fraction as a sum of partial fractions with unknown constants (A, B, C, etc.) in the numerators. The form depends on the type of factors in the denominator.

Example: P(x)/((2x + 1)(x - 3)) = A/(2x + 1) + B/(x - 3)

STEP 4: Eliminate Fractions

Multiply both sides by the common denominator (the original denominator) to eliminate all fractions. This creates an identity that must be true for all values of x.

Example: 3x + 5 = A(x - 3) + B(2x + 1)

STEP 5: Find Constants (Method 1 - Substitution)

Substitute strategic values of x (usually the roots of the denominator factors) to create simple equations for A, B, C, etc.

Let x = 3: 3(3) + 5 = A(0) + B(7) → B = 2
Let x = -1/2: 3(-1/2) + 5 = A(-7/2) + B(0) → A = -1

STEP 5: Find Constants (Method 2 - Comparing Coefficients)

Alternatively, expand the right side and compare coefficients of like terms (x², x, constants) on both sides to form simultaneous equations.

Coefficient of x: 3 = A + 2B
Constant term: 5 = -3A + B

STEP 6: Write Final Answer

Substitute the values of constants back into the partial fractions form. Always verify by combining the fractions to check you get the original expression.

🎓 Pro Strategy: The substitution method (choosing roots) is usually faster than comparing coefficients, especially for distinct linear factors. Use comparing coefficients when substitution doesn't provide enough equations.

📐 Case 1: Distinct Linear Factors

📝 Worked Example 1

Express (7x + 13)/(x² + x - 6) in partial fractions.

Step 1: Factorize denominator

x² + x - 6 = (x + 3)(x - 2)

Step 2: Set up partial fractions

(7x + 13)/((x + 3)(x - 2)) = A/(x + 3) + B/(x - 2)

Step 3: Multiply by denominator

7x + 13 = A(x - 2) + B(x + 3)

Step 4: Substitute x = 2 (root of x - 2)

7(2) + 13 = A(0) + B(5)
27 = 5B
B = 27/5

Step 5: Substitute x = -3 (root of x + 3)

7(-3) + 13 = A(-5) + B(0)
-8 = -5A
A = 8/5

Final Answer:

(7x + 13)/(x² + x - 6) = (8/5)/(x + 3) + (27/5)/(x - 2)

📝 Worked Example 2

Decompose (5x - 1)/(x² - 4) into partial fractions.

Step 1: Factorize (difference of squares)

x² - 4 = (x - 2)(x + 2)

Step 2-3: Set up and eliminate fractions

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

5x - 1 = A(x + 2) + B(x - 2)

Step 4: Let x = 2

5(2) - 1 = A(4) + B(0)
9 = 4A → A = 9/4

Step 5: Let x = -2

5(-2) - 1 = A(0) + B(-4)
-11 = -4B → B = 11/4

Final Answer:

(5x - 1)/(x² - 4) = (9/4)/(x - 2) + (11/4)/(x + 2)

🔁 Case 2: Repeated Linear Factors

When a linear factor appears more than once (e.g., (x + 2)²), we need separate terms for each power up to the highest power.

General Form: P(x)/(ax + b)ⁿ = A₁/(ax + b) + A₂/(ax + b)² + ... + A_n/(ax + b)ⁿ

📝 Worked Example 3

Express (20x + 35)/(x + 4)² in partial fractions.

Step 1: Set up (repeated factor form)

(20x + 35)/(x + 4)² = A/(x + 4) + B/(x + 4)²

Step 2: Multiply by (x + 4)²

20x + 35 = A(x + 4) + B

Step 3: Expand

20x + 35 = Ax + 4A + B

Step 4: Compare coefficients

Coefficient of x: A = 20
Constant term: 4A + B = 35
4(20) + B = 35
B = -45

Final Answer:

(20x + 35)/(x + 4)² = 20/(x + 4) - 45/(x + 4)²

📝 Worked Example 4

Decompose (3x² - 2x + 1)/((x - 1)²(x + 2)) into partial fractions.

Step 1: Set up (mixed factors)

(3x² - 2x + 1)/((x - 1)²(x + 2)) = A/(x - 1) + B/(x - 1)² + C/(x + 2)

Step 2: Multiply by (x - 1)²(x + 2)

3x² - 2x + 1 = A(x - 1)(x + 2) + B(x + 2) + C(x - 1)²

Step 3: Let x = 1 (eliminates A and C)

3(1)² - 2(1) + 1 = B(3)
2 = 3B → B = 2/3

Step 4: Let x = -2 (eliminates A and B)

3(-2)² - 2(-2) + 1 = C(-3)²
17 = 9C → C = 17/9

Step 5: Let x = 0 (find A)

1 = A(-1)(2) + B(2) + C(1)
1 = -2A + 2(2/3) + 17/9
1 = -2A + 4/3 + 17/9
-2A = 1 - 12/9 - 17/9 = -20/9
A = 10/9

Final Answer:

= (10/9)/(x - 1) + (2/3)/(x - 1)² + (17/9)/(x + 2)

⚠️ Common Error: Students often forget to include ALL powers up to the highest power for repeated factors. For (x - 1)³, you need terms for (x - 1), (x - 1)², AND (x - 1)³.

📊 Case 3: Irreducible Quadratic Factors

When the denominator contains a quadratic factor that cannot be factored further (discriminant < 0), the numerator must be linear (Ax + B), not just a constant.

General Form: P(x)/((ax + b)(cx² + dx + e)) = A/(ax + b) + (Bx + C)/(cx² + dx + e)

📝 Worked Example 5

Express 10/((x - 1)(x² + 9)) in partial fractions.

Step 1: Check if quadratic factorizes

x² + 9 has discriminant = 0² - 4(1)(9) = -36 < 0
Therefore irreducible (cannot be factored over real numbers)

Step 2: Set up with linear numerator for quadratic

10/((x - 1)(x² + 9)) = A/(x - 1) + (Bx + C)/(x² + 9)

Step 3: Multiply by (x - 1)(x² + 9)

10 = A(x² + 9) + (Bx + C)(x - 1)

Step 4: Let x = 1

10 = A(10) + 0
A = 1

Step 5: Expand and compare coefficients

10 = Ax² + 9A + Bx² - Bx + Cx - C
10 = (A + B)x² + (-B + C)x + (9A - C)

Coefficient of x²: 0 = A + B → B = -1
Coefficient of x: 0 = -B + C → C = -1
Constant: 10 = 9A - C → 10 = 9(1) - (-1) = 10 ✓

Final Answer:

10/((x - 1)(x² + 9)) = 1/(x - 1) + (-x - 1)/(x² + 9)

or: = 1/(x - 1) - (x + 1)/(x² + 9)

💡 Integration Note: When integrating partial fractions with quadratic denominators, you'll typically need to split the numerator and use substitution for one part and inverse tan for the other. This is a common IB exam technique.

⚡ Case 4: Improper Fractions

An improper fraction has a numerator with degree ≥ denominator degree. Before decomposing, we must perform polynomial long division.

If deg(P) ≥ deg(Q): P(x)/Q(x) = Quotient + Remainder/Q(x)

Then apply partial fractions only to the proper fraction (Remainder/Q(x))

📝 Worked Example 6

Express (x³ + 2x² + 3)/(x² - 1) as partial fractions.

Step 1: Check if proper

Numerator degree = 3, Denominator degree = 2
Since 3 > 2, this is IMPROPER - must divide first!

Step 2: Polynomial long division

Dividing x³ + 2x² + 3 by x² - 1:

         x + 2

     _______________

x² - 1 | x³ + 2x² + 0x + 3

         x³ + 0x² - x

         _______________

             2x² + x + 3

             2x² + 0x - 2

             _______________

                  x + 5

Result: (x³ + 2x² + 3)/(x² - 1) = x + 2 + (x + 5)/(x² - 1)

Step 3: Partial fractions on remainder

(x + 5)/(x² - 1) = (x + 5)/((x - 1)(x + 1)) = A/(x - 1) + B/(x + 1)

Step 4: Solve for A and B

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

Let x = 1: 6 = 2A → A = 3
Let x = -1: 4 = -2B → B = -2

Final Answer:

(x³ + 2x² + 3)/(x² - 1) = x + 2 + 3/(x - 1) - 2/(x + 1)

⚠️ Exam Tip: In IB exams, if you see the numerator degree ≥ denominator degree, immediately recognize you need polynomial division. This step is often worth marks on its own!

🔗 Application: Integration with Partial Fractions

🎯 Why Partial Fractions for Integration?

Many rational functions cannot be integrated directly. Partial fractions break them into simpler forms that integrate to logarithms and inverse trigonometric functions.

Key Integration Results

∫ 1/(ax + b) dx = (1/a)ln|ax + b| + C

∫ 1/(x² + a²) dx = (1/a)arctan(x/a) + C

After partial fraction decomposition, integrate each term separately using these standard forms.

📝 Integration Example

Evaluate ∫ (7x + 13)/(x² + x - 6) dx

Step 1: Use partial fractions (from earlier example)

(7x + 13)/(x² + x - 6) = (8/5)/(x + 3) + (27/5)/(x - 2)

Step 2: Integrate each term

∫ [(8/5)/(x + 3) + (27/5)/(x - 2)] dx

= (8/5)ln|x + 3| + (27/5)ln|x - 2| + C

Step 3: Simplify using log laws (optional)

= (1/5)ln|x + 3|⁸ + (1/5)ln|x - 2|²⁷ + C

= (1/5)ln|(x + 3)⁸(x - 2)²⁷| + C

✍️ Interactive Practice Problems

Test your understanding with these IB-style problems. Enter your answer and check immediately!

0% Complete

Problem 1: Distinct Linear Factors

Express (8x + 2)/(x² - 1) in partial fractions.
Give your answer in the form: A/(x - 1) + B/(x + 1)

Enter A and B:

📖 Complete Solution

Step 1: Factorize: x² - 1 = (x - 1)(x + 1)

Step 2: Set up: (8x + 2)/((x - 1)(x + 1)) = A/(x - 1) + B/(x + 1)

Step 3: Multiply: 8x + 2 = A(x + 1) + B(x - 1)

Step 4: Let x = 1: 10 = 2A → A = 5

Step 5: Let x = -1: -6 = -2B → B = 3

Answer: (8x + 2)/(x² - 1) = 5/(x - 1) + 3/(x + 1)

Problem 2: Repeated Linear Factors

Express (3x + 1)/(x + 1)² in partial fractions.
Give your answer in the form: A/(x + 1) + B/(x + 1)²

Enter A and B:

📖 Complete Solution

Step 1: Set up: (3x + 1)/(x + 1)² = A/(x + 1) + B/(x + 1)²

Step 2: Multiply by (x + 1)²: 3x + 1 = A(x + 1) + B

Step 3: Expand: 3x + 1 = Ax + A + B

Step 4: Compare coefficients:
Coefficient of x: A = 3
Constant: A + B = 1 → 3 + B = 1 → B = -2

Answer: (3x + 1)/(x + 1)² = 3/(x + 1) - 2/(x + 1)²

Problem 3: Three Linear Factors

Express (2x² + x + 1)/((x - 1)(x + 1)(x + 2)) in partial fractions.
Give your answer in the form: A/(x - 1) + B/(x + 1) + C/(x + 2)

Enter A, B, and C:

📖 Complete Solution

Step 1: Set up: = A/(x - 1) + B/(x + 1) + C/(x + 2)

Step 2: Multiply by denominator:
2x² + x + 1 = A(x + 1)(x + 2) + B(x - 1)(x + 2) + C(x - 1)(x + 1)

Step 3: Let x = 1:
2(1) + 1 + 1 = A(2)(3) → 4 = 6A → A = 2/3

Step 4: Let x = -1:
2(1) - 1 + 1 = B(-2)(1) → 2 = -2B → B = -1

Step 5: Let x = -2:
2(4) - 2 + 1 = C(-3)(-1) → 7 = 3C → C = 7/3

Answer: = (2/3)/(x - 1) - 1/(x + 1) + (7/3)/(x + 2)

Problem 4: Mixed Factors Challenge

Express (x² + 3x + 5)/((x + 1)²(x - 2)) in partial fractions.
Give your answer in the form: A/(x + 1) + B/(x + 1)² + C/(x - 2)

Enter A, B, and C:

📖 Complete Solution

Step 1: Set up: = A/(x + 1) + B/(x + 1)² + C/(x - 2)

Step 2: Multiply:
x² + 3x + 5 = A(x + 1)(x - 2) + B(x - 2) + C(x + 1)²

Step 3: Let x = -1:
1 - 3 + 5 = B(-3) → 3 = -3B → B = -1

Step 4: Let x = 2:
4 + 6 + 5 = C(9) → 15 = 9C → C = 5/3

Step 5: Let x = 0:
5 = A(1)(-2) + B(-2) + C(1)
5 = -2A - 2(-1) + 5/3
5 = -2A + 2 + 5/3 → -2A = 5 - 2 - 5/3 = 4/3
A = -2/3

Answer: = (-2/3)/(x + 1) - 1/(x + 1)² + (5/3)/(x - 2)

⚠️ Common Mistakes to Avoid

❌ Mistake 1: Wrong Form for Repeated Factors

Wrong: P(x)/(x + 2)² = A/(x + 2)²

Correct: P(x)/(x + 2)² = A/(x + 2) + B/(x + 2)²

❌ Mistake 2: Constant for Quadratic Factors

Wrong: P(x)/(x² + 4) = A/(x² + 4)

Correct: P(x)/((x - 1)(x² + 4)) = A/(x - 1) + (Bx + C)/(x² + 4)

❌ Mistake 3: Forgetting to Factor

Don't decompose before factoring denominator completely

Always factor first: x² - 4 = (x - 2)(x + 2)

❌ Mistake 4: Improper Fractions

Decomposing directly when deg(numerator) ≥ deg(denominator)

Must perform polynomial division first!

🎓 IB Exam Strategy & Tips

📝 Exam Technique

Show All Steps: IB examiners award method marks. Always show factorization, setup, and calculations clearly
Check Your Work: Quickly verify by combining your partial fractions back to the original - catches arithmetic errors
Time Management: Partial fraction questions are typically 4-8 marks and should take 5-10 minutes
Use Calculator Wisely: GDC can verify your final answer by evaluating both forms at a test value
Common Point Values: M1 for form, M1 for eliminating fractions, A1 for each constant

🎯 What Examiners Look For

Correct Form: Setting up the right partial fractions structure (M1 mark)
Valid Method: Either substitution or coefficient comparison consistently applied
Accurate Arithmetic: Correct calculation of constants (A1 marks)
Simplified Answer: Final answer with fractions in lowest terms
Follow-Through: Integration questions may allow marks even if partial fractions have errors

🔑 Key Formula Sheet Reminder: The IB formula booklet does NOT include partial fractions formulas or methods. You must know the standard forms and procedures from memory. Practice until these become automatic!

📌 Quick Reference Summary

Distinct Linear: (ax + b)(cx + d) → A/(ax + b) + B/(cx + d)
Repeated Linear: (ax + b)² → A/(ax + b) + B/(ax + b)²
Irreducible Quadratic: (ax + b)(cx² + dx + e) → A/(ax + b) + (Bx + C)/(cx² + dx + e)
Improper Fractions: Perform polynomial division first
Solving Methods: Substitution (faster) or comparing coefficients (systematic)
Applications: Integration of rational functions, differential equations

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Before you have learnt how to combine together different fractions to bring them under one denominator. However, sometimes you are required to do the opposite: split a fraction into distinct terms. In IB you will only be asked to split up fractions with two distinct linear terms in the denominator.

Solving partial fractions problems

Determine which linear terms make up the denominator

x² + x − 2 = (x − 1)(x + 2)

2. Equate the fraction to sum of two fractions with unknown constants as numerators and the linear terms as denominators

3. Multiply by the linear terms on both sides and determine the constant terms

4. Plug in constant terms into the original equation

Frequently Asked Questions: Partial Fractions

What is Partial Fraction Decomposition?

Partial fraction decomposition is a mathematical technique used to break down a rational expression (a fraction where the numerator and denominator are polynomials) into a sum of simpler fractions, called partial fractions.

Why is Partial Fraction Decomposition used?

Its primary use is in calculus, specifically for integrating rational functions. Breaking down a complex rational function into simpler partial fractions makes the integration process much easier, as the resulting terms can often be integrated using basic rules (like logarithms or arctangents).

When should I use Partial Fraction Decomposition?

You use partial fraction decomposition when you need to integrate a rational function (a polynomial divided by another polynomial) where the degree of the numerator is less than the degree of the denominator (a proper fraction). If the degree of the numerator is greater than or equal to the denominator (an improper fraction), you first perform polynomial long division.

How do you do Partial Fraction Decomposition?

The general steps are:

Check if the fraction is proper. If not, divide the polynomials first.
Factor the denominator completely into linear factors (ax+b) and irreducible quadratic factors (ax²+bx+c where b²-4ac < 0).
Set up the form of the partial fraction decomposition based on the factors.
Solve for the unknown constants (like A, B, C) in the numerator(s) using methods such as equating coefficients or substituting roots of the denominator.

How do I set up the partial fractions for different types of factors?

The setup depends on the factors of the denominator:

Linear Distinct Factors (ax+b): For each distinct linear factor, use a term like A / (ax+b).
Repeated Linear Factors ((ax+b)ⁿ): For a factor repeated n times, use a sum of n terms: A₁ / (ax+b) + A₂ / (ax+b)² + ... + An / (ax+b)ⁿ.
Irreducible Quadratic Factors (ax²+bx+c): For each distinct irreducible quadratic factor, use a term like (Ax+B) / (ax²+bx+c).
Repeated Irreducible Quadratic Factors ((ax²+bx+c)ⁿ): For a factor repeated n times, use a sum of n terms: (A₁x+B₁) / (ax²+bx+c) + (A₂x+B₂) / (ax²+bx+c)² + ... + (Anx+Bn) / (ax²+bx+c)ⁿ.

How do I solve for the unknown constants (A, B, etc.)?

After setting up the form, multiply both sides of the equation by the original denominator to clear the denominators. Then you can solve for the constants using two main methods:

Method of Equating Coefficients: Expand the right side of the equation and group terms by powers of x (x², x, constant). Equate the coefficients of corresponding powers of x on both sides of the equation to get a system of linear equations. Solve this system for the constants.
Method of Substitution (Heaviside Cover-Up Method): Substitute the roots of the original denominator (the values of x that make the denominator zero) into the equation (after clearing denominators). This can directly give you the values of some constants, especially for distinct linear factors. For irreducible quadratic factors or repeated roots, a combination of methods might be needed.

How do I integrate using Partial Fractions?

Once you have decomposed the original rational function into a sum of simpler partial fractions, you integrate each partial fraction separately. The integration of these simpler forms is standard:

Terms like A / (ax+b) integrate to (A/a) ln|ax+b| + C.
Terms like B / (ax+b)ⁿ (n ≠ 1) integrate using the power rule: B ∫ (ax+b)-n dx.
Terms like (Ax+B) / (ax²+bx+c) often require splitting into two integrals: one that integrates to a logarithm (related to the derivative of the denominator) and one that integrates to an arctangent.

Sum the results of integrating each partial fraction to get the final integral of the original rational function.

What if the rational function is improper (degree of numerator ≥ degree of denominator)?

If the degree of the numerator is greater than or equal to the degree of the denominator, you must perform polynomial long division first. This will result in a polynomial plus a proper rational fraction. You then apply partial fraction decomposition only to the proper rational fraction part. The integral will be the integral of the polynomial plus the integral of the partial fractions.