Partial Fractions in Integration — AHL 5.10 Partial fractions in integration (HL link) - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

When you first learned fractions in arithmetic, you discovered that adding 1/2 + 1/3 required a common denominator, giving you 5/6. Partial fraction decomposition is the reverse process: starting with a single complicated fraction and splitting it back into the simpler pieces that were originally added together. This algebraic technique became essential to calculus because many integrals involve rational expressions — fractions where both the numerator and denominator are polynomials. Without a way to break them apart, these integrals would be almost impossible to evaluate by hand.

1655 • John Wallis & Early Decompositions

English mathematician John Wallis explored ways to decompose rational expressions while studying infinite series and the area under curves.

1702 • Leibniz–Bernoulli Debate

Gottfried Leibniz and Johann Bernoulli debated whether every rational function could be decomposed into partial fractions, driving the technique into mainstream calculus.

1740 • Euler's Systematic Approach

Leonhard Euler formalized partial fraction rules for all polynomial denominators, including repeated and irreducible quadratic factors, making the method a standard integration tool.

Modern • IB & University Curriculum Staple

Today, partial fractions appear in IB HL Mathematics, AP Calculus BC, and virtually every university calculus course as a core integration strategy.

The central question that partial fractions answers is straightforward: how do you integrate a fraction like (3x + 5) / (x² + 3x + 2) when no single substitution or basic rule applies? By splitting it into pieces whose integrals you already know, you transform an intimidating problem into routine work.

SECTION 2

Core Principles & Definitions

Before diving into the mechanics, you need to understand several foundational ideas that make partial fractions work. A rational function is any expression of the form P(x)/Q(x), where both P and Q are polynomials. The technique of partial fractions only applies when the degree of P is strictly less than the degree of Q — that is, the fraction must be proper. If the fraction is improper, you must first perform polynomial long division to extract a polynomial part and a proper remainder.

Proper vs. Improper Fractions

If deg(P) < deg(Q), the fraction is proper and ready for decomposition. If not, divide first to obtain a polynomial plus a proper fraction.

Factor the Denominator

Every polynomial with real coefficients factors into linear factors (ax + b) and irreducible quadratic factors (ax² + bx + c where b² − 4ac < 0). This factoring is the foundation of the decomposition.

One Fraction per Factor

Each distinct linear factor contributes a term A/(ax + b). Each irreducible quadratic factor contributes (Ax + B)/(ax² + bx + c). Repeated factors require additional terms for each power.

Solve for Constants

Multiply both sides by the full denominator, then use substitution or equating coefficients to find the unknown constants A, B, C, etc.

Integrate Each Piece

Each resulting fraction integrates using ln|ax + b| for linear denominators, or arctan and logarithm combinations for quadratic denominators.

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KEY TAKEAWAY

Think of partial fractions like un-mixing a smoothie. If someone blended orange juice and apple juice together, it would be hard to drink just the orange juice. But if you knew the recipe — the original separate ingredients — you could work with each one individually. Partial fractions reverse the blending of fractions so you can integrate each ingredient on its own.

SECTION 3

Visual Explanation

The Decomposition Flowchart

This flowchart shows the complete decision process: check if the fraction is proper, factor the denominator, assign the correct partial fraction form, and integrate each piece using logarithmic or inverse tangent antiderivatives.

The flowchart above captures the full algorithm. Notice that every path leads to integrals you have already mastered: natural logarithm for linear denominators and combinations of ln and arctan for irreducible quadratic denominators. The power of partial fractions is that it reduces unfamiliar integrals to familiar ones.

SECTION 4

Mathematical Framework

In the IB HL syllabus (AHL 5.10), you are expected to handle denominators that factor into distinct linear factors. While repeated and quadratic factors exist in broader mathematics, the IB focuses on the case where the denominator splits into non-repeating linear terms. Here is the formal setup.

GENERAL DECOMPOSITION — DISTINCT LINEAR FACTORS

P(x) / [(x − r₁)(x − r₂)···(x − rₙ)] = A₁/(x − r₁) + A₂/(x − r₂) + ··· + Aₙ/(x − rₙ)

P(x) is a polynomial of degree less than n. Each r_i is a distinct root of the denominator, and each A_i is a constant to be determined.

KEY INTEGRATION RESULT

∫ A/(x − r) dx = A · ln|x − r| + C

This is the fundamental antiderivative that makes partial fractions useful. Once you decompose, every term integrates to a logarithm.

TWO-FACTOR CASE (MOST COMMON IN IB)

(px + q) / [(x − a)(x − b)] = A/(x − a) + B/(x − b)

Multiply both sides by (x − a)(x − b) to get px + q = A(x − b) + B(x − a). Substitute x = a and x = b to find A and B directly.

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Cover-Up Method (Quick Trick)

To find the constant above a particular factor, cover up that factor in the original fraction and substitute the root of that factor into what remains. For example, to find A, cover (x − a) in the original fraction and substitute x = a into the rest. This is often called the Heaviside cover-up method and saves considerable algebra.

SECTION 5

Detailed Breakdown of Cases

Different denominator structures require different decomposition templates. The table below summarizes the main cases you will encounter, with the IB HL focus highlighted. Understanding which template to use is half the battle.

Denominator Type	Partial Fraction Form	IB HL Syllabus?
Distinct linear: (x − a)(x − b)	A/(x − a) + B/(x − b)	Yes — core focus
Three distinct linear: (x − a)(x − b)(x − c)	A/(x − a) + B/(x − b) + C/(x − c)	Yes
Repeated linear: (x − a)²	A/(x − a) + B/(x − a)²	Awareness only
Irreducible quadratic: (x² + bx + c)	(Ax + B)/(x² + bx + c)	Not required

Summary of partial fraction forms by denominator type

The solid violet curve is the original rational function 1/[(x − 1)(x − 3)]. The dashed green and pink curves are its two partial fraction components: −½/(x − 1) and ½/(x − 3). Notice how the two simpler curves add up to produce the original curve at every point (except at the vertical asymptotes x = 1 and x = 3).

The graph above illustrates the core idea visually. The original function, shown in violet, is the sum of the two simpler fractions shown in green and pink. Integrating each dashed curve individually is straightforward because each one has the form A/(x − r), whose antiderivative is simply A · ln|x − r| + C.

SECTION 6

Worked Example

Let's walk through a complete IB-style problem step by step. We will integrate ∫ (5x + 1) / (x² − x − 2) dx from start to finish.

Evaluate ∫ (5x + 1)/(x² − x − 2) dx

1
Step 1 — Check if the fraction is proper

The numerator 5x + 1 has degree 1, and the denominator x² − x − 2 has degree 2. Since 1 < 2, the fraction is proper, so no long division is needed.

Proper fraction ✓
2
Step 2 — Factor the denominator

We need two numbers that multiply to −2 and add to −1. Those numbers are −2 and +1. So x² − x − 2 = (x − 2)(x + 1).

x² − x − 2 = (x − 2)(x + 1)
3
Step 3 — Set up the partial fraction decomposition

Write (5x + 1)/[(x − 2)(x + 1)] = A/(x − 2) + B/(x + 1). Now multiply both sides by (x − 2)(x + 1) to clear the denominators: 5x + 1 = A(x + 1) + B(x − 2).

5x + 1 = A(x + 1) + B(x − 2)
4
Step 4 — Solve for A and B using substitution

Substitute x = 2: 5(2) + 1 = A(2 + 1) + B(0), so 11 = 3A, giving A = 11/3. Substitute x = −1: 5(−1) + 1 = A(0) + B(−1 − 2), so −4 = −3B, giving B = 4/3.

A = 11/3, B = 4/3
5
Step 5 — Write the decomposition

The original fraction becomes (11/3)/(x − 2) + (4/3)/(x + 1). This is equivalent to (11/3) · 1/(x − 2) + (4/3) · 1/(x + 1).

(5x + 1)/[(x − 2)(x + 1)] = (11/3)·1/(x − 2) + (4/3)·1/(x + 1)
6
Step 6 — Integrate each term

∫ (11/3)·1/(x − 2) dx = (11/3) ln|x − 2| + C₁ and ∫ (4/3)·1/(x + 1) dx = (4/3) ln|x + 1| + C₂. Combining these with a single constant C gives the final answer.

∫ (5x + 1)/(x² − x − 2) dx = (11/3) ln|x − 2| + (4/3) ln|x + 1| + C

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Verification Tip

You can always check your decomposition by adding the partial fractions back together. If (11/3)/(x − 2) + (4/3)/(x + 1) simplifies back to (5x + 1)/[(x − 2)(x + 1)], you know A and B are correct before you even start integrating.

SECTION 7

Strengths, Limitations & Comparisons

Partial fractions is a powerful technique, but it is not the right tool for every integral. Knowing when to use it — and when another method is more efficient — is just as important as knowing how to do it.

Strengths	Limitations
Works systematically for any rational function with factorable denominator.	Only applies to rational functions (polynomial over polynomial).
Reduces complex integrals to standard forms (ln and arctan).	Requires the denominator to be fully factored — this step can itself be difficult.
The algebra is routine once you learn the pattern — no clever substitutions needed.	Can be algebraically heavy with three or more factors.
Integrates nicely with other techniques (e.g., long division for improper fractions).	Does not help with irrational or transcendental integrands like √x or sin x.

When partial fractions works well vs. when to choose a different method

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CHOOSING YOUR TOOL

Think of your integration techniques like tools in a toolbox. Substitution is your screwdriver — versatile and often the first thing you try. Integration by parts is a wrench for products of different function types. Partial fractions is a specialized socket wrench: when you see a rational function with a factorable denominator, it is exactly the right tool. Recognizing the form of the integrand tells you which tool to grab.

SECTION 8

Connection to Advanced Theory

The IB HL syllabus focuses on distinct linear factors, but the full theory of partial fractions extends to cases you may encounter in university mathematics, engineering, and physics. Here is a brief preview of how the technique grows.

Feature	IB HL Level (AHL 5.10)	University / Advanced Level
Denominator factors	Distinct linear factors only	Repeated linear, irreducible quadratics, and combinations
Resulting integrals	ln\|x − r\| forms	Also arctan, power rule for repeated factors, completing the square
Applications	Evaluating definite and indefinite integrals	Laplace transforms, control systems, signal processing, differential equations
Coefficients	Found by substitution or equating coefficients	Same techniques plus residue calculus in complex analysis

IB HL scope vs. university-level extensions

If you continue studying mathematics or engineering at university, you will find that partial fractions reappear in unexpected places. In electrical engineering, for instance, the Laplace transform converts differential equations into algebraic equations in a new variable, and partial fractions is the key step in converting the solution back to the time domain. Mastering the technique now builds a foundation for these advanced applications.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why partial fraction decomposition requires the fraction to be proper (i.e., the degree of the numerator must be strictly less than the degree of the denominator). What must you do if the fraction is improper?

Show answer

Partial fraction decomposition works by expressing a rational function as a sum of simpler fractions, each with a denominator that is one factor of the original denominator. This is only possible when the numerator has a lower degree than the denominator because a higher-degree numerator would mean the fraction contains a hidden polynomial part that cannot be expressed in the A/(x − r) form. If the fraction is improper, you must perform polynomial long division first to separate the polynomial quotient from the proper remainder. You then apply partial fractions only to the proper remainder. This ensures every piece of the decomposition fits the required template.

Question 2: BASIC CALCULATION

Decompose (3x + 7)/[(x + 1)(x + 3)] into partial fractions. That is, find constants A and B such that (3x + 7)/[(x + 1)(x + 3)] = A/(x + 1) + B/(x + 3).

Show answer

Multiply both sides by (x + 1)(x + 3): 3x + 7 = A(x + 3) + B(x + 1). Substitute x = −1: 3(−1) + 7 = A(−1 + 3) + 0, so 4 = 2A, giving A = 2. Substitute x = −3: 3(−3) + 7 = 0 + B(−3 + 1), so −2 = −2B, giving B = 1. Therefore (3x + 7)/[(x + 1)(x + 3)] = 2/(x + 1) + 1/(x + 3). You can verify by combining the right side over a common denominator: [2(x + 3) + 1(x + 1)] / [(x + 1)(x + 3)] = (2x + 6 + x + 1)/[(x + 1)(x + 3)] = (3x + 7)/[(x + 1)(x + 3)] ✓.

Question 3: INTERMEDIATE

Evaluate the integral ∫ (2x + 3)/(x² + x − 6) dx. Show all steps including factoring, decomposition, and integration.

Show answer

First, factor the denominator: x² + x − 6 = (x + 3)(x − 2). Set up the decomposition: (2x + 3)/[(x + 3)(x − 2)] = A/(x + 3) + B/(x − 2). Multiply through: 2x + 3 = A(x − 2) + B(x + 3). Substitute x = 2: 7 = 5B, so B = 7/5. Substitute x = −3: −3 = −5A, so A = 3/5. Now integrate: ∫ (3/5)·1/(x + 3) dx + ∫ (7/5)·1/(x − 2) dx = (3/5) ln|x + 3| + (7/5) ln|x − 2| + C. The key was recognizing the factorable quadratic denominator and carefully solving for both constants before integrating.

Question 4: APPLIED

A population model gives the rate of growth as dP/dt = 120/[(P − 10)(40 − P)]. Use partial fractions to find ∫ (P − 10)(40 − P) dP / 120, which arises when separating variables. Decompose 1/[(P − 10)(40 − P)] into partial fractions and integrate with respect to P.

Show answer

Write 1/[(P − 10)(40 − P)] = A/(P − 10) + B/(40 − P). Multiply through: 1 = A(40 − P) + B(P − 10). Substitute P = 10: 1 = 30A, so A = 1/30. Substitute P = 40: 1 = 30B, so B = 1/30. The decomposition is (1/30)·[1/(P − 10) + 1/(40 − P)]. Integrating: ∫ (1/30)/(P − 10) dP + ∫ (1/30)/(40 − P) dP = (1/30) ln|P − 10| − (1/30) ln|40 − P| + C = (1/30) ln|(P − 10)/(40 − P)| + C. Note the negative sign in the second integral comes from the chain rule since d/dP(40 − P) = −1. This result is the standard logistic-style integral that models constrained population growth.

Question 5: CRITICAL THINKING

Consider the integral ∫ x²/(x² − 1) dx. This is an improper rational function. Show how to use long division followed by partial fractions to evaluate the integral completely.

Show answer

Since the numerator and denominator both have degree 2, the fraction is improper. Perform long division: x²/(x² − 1) = 1 + 1/(x² − 1), because x² = 1·(x² − 1) + 1. Now factor x² − 1 = (x − 1)(x + 1) and decompose 1/[(x − 1)(x + 1)] = A/(x − 1) + B/(x + 1). Multiply through: 1 = A(x + 1) + B(x − 1). Substitute x = 1: 1 = 2A, so A = 1/2. Substitute x = −1: 1 = −2B, so B = −1/2. Therefore x²/(x² − 1) = 1 + (1/2)/(x − 1) − (1/2)/(x + 1). Integrating: ∫ 1 dx + (1/2) ∫ 1/(x − 1) dx − (1/2) ∫ 1/(x + 1) dx = x + (1/2) ln|x − 1| − (1/2) ln|x + 1| + C = x + (1/2) ln|(x − 1)/(x + 1)| + C. This problem synthesizes long division and partial fractions, demonstrating that the two techniques work hand-in-hand.

SUMMARY

Lesson Summary

Summary

Partial fraction decomposition is an algebraic technique that breaks a complicated rational function into a sum of simpler fractions, each with a single factor in the denominator. The method requires the fraction to be proper (numerator degree less than denominator degree); if it is not, you first perform polynomial long division. After factoring the denominator into distinct linear factors, you assign unknown constants A, B, C, … to each factor, then solve for these constants using substitution or equating coefficients.

Once decomposed, each term integrates to a natural logarithm: ∫ A/(x − r) dx = A · ln|x − r| + C. The IB HL syllabus (AHL 5.10) focuses on distinct linear factors, while university courses extend the technique to repeated factors and irreducible quadratics. Mastering partial fractions gives you a reliable, systematic method for integrating rational expressions and prepares you for advanced applications in differential equations, Laplace transforms, and engineering.