How to Solve Partial Fraction Decomposition: Complete Step-by-Step Guide
Partial fraction decomposition is a technique for breaking a rational expression into a sum of simpler fractions. It appears in algebra, precalculus, and calculus — especially when integrating rational functions. If you have ever tried to integrate something like (3x + 5) / ((x + 1)(x + 2)) and felt stuck, this guide covers the exact steps you need. Every case type — distinct linear factors, repeated factors, and irreducible quadratic factors — is shown with fully worked examples and a verification step.
Contents
- 01What Is Partial Fraction Decomposition?
- 02When to Use Partial Fraction Decomposition
- 03Worked Example 1: Distinct Linear Factors
- 04Worked Example 2: Repeated Linear Factors
- 05Worked Example 3: Irreducible Quadratic Factors
- 06Common Mistakes and How to Avoid Them
- 07Practice Problems with Solutions
- 08Tips for Faster Partial Fraction Decomposition
- 09Partial Fraction Decomposition in Calculus Integration
- 10Frequently Asked Questions
What Is Partial Fraction Decomposition?
Partial fraction decomposition (PFD) is the reverse process of adding fractions. When you add 2/(x + 1) + 3/(x + 2), you get a single combined rational expression. PFD works backward: you start with the combined fraction and split it into simpler parts. The technique applies to proper rational functions — fractions where the degree of the numerator is strictly less than the degree of the denominator. If the numerator degree is equal to or greater than the denominator degree, you must perform polynomial long division first to reduce it before decomposing. The resulting simpler fractions are called partial fractions, and they are significantly easier to integrate, simplify, or work with in differential equations.
Partial fraction decomposition converts one complicated fraction into a sum of simpler ones — making integration and algebraic manipulation far more manageable.
When to Use Partial Fraction Decomposition
You will encounter partial fraction decomposition in three main contexts: integrating rational functions in calculus, simplifying complex algebraic expressions, and solving differential equations using Laplace transforms. The setup depends entirely on the types of factors in the denominator. There are three cases: distinct linear factors such as (x + 1)(x − 3), repeated linear factors such as (x − 2)², and irreducible quadratic factors such as (x² + 4) that cannot be factored over the real numbers. Each case follows a specific template for writing the partial fractions. Recognizing which case you are dealing with before you start is half the work.
1. Step 1 — Check if the fraction is proper
Compare the degree of the numerator to the degree of the denominator. If the numerator degree is strictly less than the denominator degree, the fraction is proper and you can proceed. If the numerator degree is greater than or equal to the denominator degree, the fraction is improper — perform polynomial long division first to produce a polynomial plus a proper remainder fraction, then decompose only the remainder.
2. Step 2 — Factor the denominator completely
Factor the denominator into linear factors (ax + b) and irreducible quadratic factors (ax² + bx + c) over the real numbers. For example, x³ − x = x(x − 1)(x + 1). A quadratic factor is irreducible when its discriminant b² − 4ac is negative — meaning it has no real roots and cannot be split further.
3. Step 3 — Write the partial fraction template
Each distinct linear factor (ax + b) gets a constant numerator: A/(ax + b). Each repeated linear factor (ax + b)ⁿ gets n separate terms: A/(ax + b) + B/(ax + b)² + ... up to the nth power. Each irreducible quadratic factor (ax² + bx + c) gets a linear numerator: (Ax + B)/(ax² + bx + c).
Worked Example 1: Distinct Linear Factors
The simplest and most common case involves a denominator with distinct (non-repeating) linear factors. Consider the rational expression (5x + 1) / ((x + 1)(x − 2)). The denominator has two distinct linear factors, and the numerator degree (1) is less than the denominator degree (2), so no long division is needed. The partial fraction template is A/(x + 1) + B/(x − 2). You multiply both sides by (x + 1)(x − 2) to eliminate the denominators, producing a polynomial identity. Substituting the roots of the denominator — x = −1 and x = 2 — into that identity lets you solve for A and B directly without expanding everything.
1. Write the template and multiply through
Set up: (5x + 1)/((x + 1)(x − 2)) = A/(x + 1) + B/(x − 2). Multiply both sides by (x + 1)(x − 2): 5x + 1 = A(x − 2) + B(x + 1).
2. Substitute x = 2 to find B
Plug in x = 2: 5(2) + 1 = A(2 − 2) + B(2 + 1) → 11 = 0 + 3B → B = 11/3.
3. Substitute x = −1 to find A
Plug in x = −1: 5(−1) + 1 = A(−1 − 2) + B(0) → −4 = −3A → A = 4/3.
4. Write the final decomposition
The partial fraction decomposition is: (5x + 1)/((x + 1)(x − 2)) = 4/(3(x + 1)) + 11/(3(x − 2)).
5. Verify by recombining
Add the two fractions: [4(x − 2) + 11(x + 1)] / (3(x + 1)(x − 2)) = [4x − 8 + 11x + 11] / (3(x + 1)(x − 2)) = (15x + 3) / (3(x + 1)(x − 2)) = (5x + 1)/((x + 1)(x − 2)) ✓
Always verify your partial fractions by recombining — if you get back the original expression, the decomposition is correct.
Worked Example 2: Repeated Linear Factors
When a linear factor appears more than once in the denominator, each power needs its own separate term. Consider (2x + 3) / ((x − 1)²(x + 2)). Here (x − 1) is a repeated factor with multiplicity 2, and (x + 2) is a distinct factor. The partial fraction template must include three terms: A/(x − 1) + B/(x − 1)² + C/(x + 2). The repeated factor (x − 1)² requires a term for each power — first and second. This pattern extends to higher multiplicities: a factor repeated n times requires n separate terms. A common error is including only the highest power and omitting the lower-power terms, which leads to an unsolvable system.
1. Set up the template and multiply through
Write: (2x + 3)/((x − 1)²(x + 2)) = A/(x − 1) + B/(x − 1)² + C/(x + 2). Multiply both sides by (x − 1)²(x + 2): 2x + 3 = A(x − 1)(x + 2) + B(x + 2) + C(x − 1)².
2. Substitute x = 1 to find B
Let x = 1: 2(1) + 3 = A(0)(3) + B(3) + C(0)² → 5 = 3B → B = 5/3.
3. Substitute x = −2 to find C
Let x = −2: 2(−2) + 3 = A(−3)(0) + B(0) + C(−3)² → −1 = 9C → C = −1/9.
4. Compare x² coefficients to find A
Expand the right side and collect x² terms: A·x² + B·0 + C·x². Comparing x² coefficients on both sides: 0 = A + C → 0 = A − 1/9 → A = 1/9. You can confirm this is consistent by checking the x and constant coefficients as well.
5. Write the final decomposition
The partial fraction decomposition is: (2x + 3)/((x − 1)²(x + 2)) = 1/(9(x − 1)) + 5/(3(x − 1)²) − 1/(9(x + 2)).
Worked Example 3: Irreducible Quadratic Factors
When the denominator contains a quadratic factor that cannot be factored over the real numbers — meaning its discriminant b² − 4ac < 0 — the corresponding partial fraction must have a linear numerator, not just a constant. Consider (3x² + 2x + 1) / ((x − 1)(x² + x + 1)). The discriminant of x² + x + 1 is 1² − 4(1)(1) = −3 < 0, confirming it is irreducible. The partial fraction template is A/(x − 1) + (Bx + C)/(x² + x + 1). The numerator for the quadratic factor is the linear expression Bx + C, which introduces two unknowns instead of one. This is why irreducible quadratic factors require more work — you cannot isolate B and C through substitution alone and must compare polynomial coefficients.
1. Set up the template and multiply through
Write: (3x² + 2x + 1)/((x − 1)(x² + x + 1)) = A/(x − 1) + (Bx + C)/(x² + x + 1). Multiply both sides by (x − 1)(x² + x + 1): 3x² + 2x + 1 = A(x² + x + 1) + (Bx + C)(x − 1).
2. Substitute x = 1 to find A
Let x = 1: 3 + 2 + 1 = A(1 + 1 + 1) + (B + C)(0) → 6 = 3A → A = 2.
3. Expand and compare coefficients for B and C
Expand the right side: 2(x² + x + 1) + (Bx + C)(x − 1) = 2x² + 2x + 2 + Bx² − Bx + Cx − C. Grouping: (2 + B)x² + (2 − B + C)x + (2 − C). Comparing x² coefficients: 3 = 2 + B → B = 1. Comparing constant terms: 1 = 2 − C → C = 1.
4. Write the final decomposition
The partial fraction decomposition is: (3x² + 2x + 1)/((x − 1)(x² + x + 1)) = 2/(x − 1) + (x + 1)/(x² + x + 1). Verify: [2(x² + x + 1) + (x + 1)(x − 1)]/((x − 1)(x² + x + 1)) = [2x² + 2x + 2 + x² − 1]/((x − 1)(x² + x + 1)) = (3x² + 2x + 1)/((x − 1)(x² + x + 1)) ✓
For irreducible quadratic factors, the numerator of the partial fraction must be linear (Ax + B), not just a constant — using only a constant will give you an incorrect result.
Common Mistakes and How to Avoid Them
Partial fraction decomposition has several predictable traps. Students often set up the wrong template, make algebra errors when finding coefficients, or forget to check whether the fraction is proper before starting. Knowing these mistakes in advance prevents them on exams, where a template error invalidates the entire calculation.
1. Mistake 1 — Using a constant numerator for a quadratic factor
Wrong: A/(x² + 4). Correct: (Ax + B)/(x² + 4). Quadratic denominators always need a linear numerator. A constant numerator gives you too few unknowns, and the resulting system will be inconsistent — meaning no valid solution exists for the constants.
2. Mistake 2 — Missing terms for repeated factors
Wrong: only A/(x − 3)² when the factor is (x − 3)². Correct: A/(x − 3) + B/(x − 3)². You need one term for every power from 1 up to the multiplicity. Omitting the lower-power terms is the most common repeated-factor error.
3. Mistake 3 — Skipping long division for improper fractions
If the numerator degree ≥ denominator degree, the fraction is improper. Example: (x³ + 2x)/(x² − 1) must be divided first. Dividing gives quotient x with remainder 3x, so (x³ + 2x)/(x² − 1) = x + 3x/(x² − 1). Only the remainder 3x/(x² − 1) gets decomposed into partial fractions.
4. Mistake 4 — Expanding everything instead of substituting roots
The substitution method — plugging in roots of the denominator — is faster and less error-prone than fully expanding and matching every coefficient. Use substitution to isolate as many constants as possible. Reserve coefficient comparison only for the unknowns that substitution cannot reach, such as A in a repeated-factor problem where the factor appears in every term.
5. Mistake 5 — Skipping the verification step
Always add your partial fractions back together and confirm you recover the original expression. This takes under a minute and catches the vast majority of errors. An incorrect decomposition leads to a wrong integral or wrong algebraic simplification — verifying first is always worth the time.
Practice Problems with Solutions
Work through these problems before looking at the solutions. The first two use distinct linear factors, the third uses a repeated factor, and the fourth involves an irreducible quadratic factor. These represent the full range of problem types you will encounter in a precalculus or calculus course.
1. Problem 1 — (7x − 3) / ((x + 2)(x − 1))
Template: A/(x + 2) + B/(x − 1). Multiply through: 7x − 3 = A(x − 1) + B(x + 2). Substitute x = 1: 4 = 3B → B = 4/3. Substitute x = −2: −17 = −3A → A = 17/3. Answer: 17/(3(x + 2)) + 4/(3(x − 1)).
2. Problem 2 — (x + 5) / (x² − x − 6)
First factor the denominator: x² − x − 6 = (x − 3)(x + 2). Template: A/(x − 3) + B/(x + 2). Multiply through: x + 5 = A(x + 2) + B(x − 3). Substitute x = 3: 8 = 5A → A = 8/5. Substitute x = −2: 3 = −5B → B = −3/5. Answer: 8/(5(x − 3)) − 3/(5(x + 2)).
3. Problem 3 — (x² + 3) / (x(x − 1)²)
Template: A/x + B/(x − 1) + C/(x − 1)². Multiply through: x² + 3 = A(x − 1)² + Bx(x − 1) + Cx. Substitute x = 0: 3 = A → A = 3. Substitute x = 1: 4 = C. Compare x² coefficients: 1 = A + B = 3 + B → B = −2. Answer: 3/x − 2/(x − 1) + 4/(x − 1)².
4. Problem 4 — (2x² + x + 4) / (x(x² + 4))
Note x² + 4 has discriminant 0 − 16 = −16 < 0, so it is irreducible. Template: A/x + (Bx + C)/(x² + 4). Multiply through: 2x² + x + 4 = A(x² + 4) + (Bx + C)x. Substitute x = 0: 4 = 4A → A = 1. Compare x² coefficients: 2 = A + B = 1 + B → B = 1. Compare x coefficients: 1 = C. Answer: 1/x + (x + 1)/(x² + 4).
Tips for Faster Partial Fraction Decomposition
Once you understand the core method, these strategies reduce time per problem — especially useful in timed exams where setting up and solving the system quickly matters.
1. Use the Heaviside cover-up method for distinct linear factors
For fractions with only distinct linear factors, you can find each constant without multiplying through. To find the coefficient for factor (x − r), cover up (x − r) in the original denominator and evaluate the remaining expression at x = r. For (5x + 1)/((x + 1)(x − 2)), the coefficient for 1/(x − 2) is found by covering (x − 2) and evaluating at x = 2: (5(2) + 1)/(2 + 1) = 11/3. Instant result — no algebra required.
2. Count your unknowns before solving
The total number of unknown constants (A, B, C, ...) must equal the degree of the denominator. For a degree-3 denominator you need exactly 3 unknowns. If you have more or fewer, your template is wrong — fix it before wasting time solving an incorrect system.
3. Mix substitution and coefficient comparison
Substitute roots of the denominator to isolate as many constants as possible — this is always the fastest path. Use coefficient comparison only for the constants that substitution cannot isolate. Do not expand and compare everything if substitution handles most of the work.
4. Learn the common denominator factoring patterns
The faster you factor the denominator, the faster you set up the correct template. Drill these: difference of squares x² − a² = (x − a)(x + a), perfect square trinomial (x ± a)², and sum/difference of cubes x³ ± a³ = (x ± a)(x² ∓ ax + a²). These cover the majority of denominators in textbook partial fraction problems.
The number of unknown constants must equal the degree of the denominator — use this as a quick sanity check before solving.
Partial Fraction Decomposition in Calculus Integration
Partial fraction decomposition is most commonly applied in calculus to evaluate integrals of rational functions. After decomposing, each partial fraction integrates using basic rules. A term A/(x − a) integrates to A · ln|x − a| + C. A repeated-factor term B/(x − a)² integrates to −B/(x − a) + C. A quadratic term (Ax + B)/(x² + k²) integrates to a combination of a natural log and an arctangent. This is why the technique is a required topic in AP Calculus BC and university calculus courses — it converts what would otherwise be very difficult integrals into straightforward ones.
1. Integration using the result from Worked Example 1
From Example 1: (5x + 1)/((x + 1)(x − 2)) = 4/(3(x + 1)) + 11/(3(x − 2)). Integrating: ∫ (5x + 1)/((x + 1)(x − 2)) dx = (4/3) · ln|x + 1| + (11/3) · ln|x − 2| + C. Without partial fraction decomposition, this integral has no direct formula — the technique reduces it to two elementary logarithm integrals.
2. Integration with a quadratic factor term
For the term (x + 1)/(x² + x + 1) from Example 3, rewrite the numerator in terms of the derivative of the denominator: d/dx(x² + x + 1) = 2x + 1. Write x + 1 = (1/2)(2x + 1) + (1/2), then split: (1/2)(2x + 1)/(x² + x + 1) + (1/2) · 1/(x² + x + 1). The first part integrates to (1/2) · ln|x² + x + 1|. The second part requires completing the square on x² + x + 1 and produces an arctangent term.
Frequently Asked Questions
These are the questions that come up most often when students first work through partial fraction decomposition problems.
1. Does partial fraction decomposition always work?
Yes, for any proper rational function with real coefficients. The method always works as long as you factor the denominator completely over the real numbers and use the correct template for each factor type. The only prerequisite is that the fraction must be proper — if it is not, divide first.
2. How do I know if a quadratic factor is irreducible?
Calculate the discriminant: b² − 4ac for the quadratic ax² + bx + c. If the discriminant is negative (< 0), the quadratic has no real roots and is irreducible over the reals. Example: x² + x + 1 has discriminant 1 − 4 = −3 < 0, so it is irreducible. Example: x² − 5x + 6 has discriminant 25 − 24 = 1 > 0, so it factors as (x − 2)(x − 3) and is not irreducible.
3. What is the difference between a proper and improper rational function?
A proper rational function has numerator degree strictly less than denominator degree. Example: (x + 1)/(x² − 1) is proper. An improper rational function has numerator degree ≥ denominator degree. Example: (x³ + 1)/(x² − 1) is improper. Only proper fractions can be directly decomposed — improper ones require polynomial long division first to extract a polynomial plus a proper remainder.
4. How many practice problems do I need before this feels natural?
Most students feel confident after 10–15 problems covering all three cases. Focus especially on repeated factors (at least 5 problems) since that is the case most often done incorrectly. The process is highly structured and algorithmic, so accuracy and speed improve quickly with focused repetition.
5. Can I use partial fractions when the denominator has complex roots?
In standard precalculus and calculus courses, you factor the denominator over the real numbers only — complex roots are left as irreducible quadratic factors. In advanced courses such as complex analysis, you can factor over the complex numbers and get simpler partial fractions with no linear numerators. Unless your course explicitly requires complex roots, stick to real factoring.
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