What Is Partial Fractions Decomposition?
Partial fractions decomposition is a technique for rewriting a complex rational expression (a fraction where both numerator and denominator are polynomials) as a sum of simpler fractions. For example, the expression 5/(x² − 1) can be decomposed into 5/2 · 1/(x − 1) − 5/2 · 1/(x + 1).
Why bother? Because the simple fractions are far easier to integrate, differentiate, or manipulate algebraically. In calculus, partial fractions is one of the essential integration techniques — it turns integrals that look impossible into straightforward logarithmic or arctangent integrals.
The idea behind partial fractions is reversing the process of adding fractions. When you add A/(x − 1) + B/(x + 1), you get a single fraction with a combined denominator. Partial fractions works backward: given the combined fraction, find the values of A and B that produce it.
When to Use Partial Fractions
Partial fractions decomposition is used in three main situations.
| # | Situation | Why Partial Fractions Helps |
|---|---|---|
| 1 | Integration | When you encounter an integral like ∫ (3x + 5)/(x² + 3x + 2) dx, you can't directly apply basic integration rules. But if you decompose the integrand into A/(x + 1) + B/(x + 2), each term integrates to a natural logarithm. This is the most common application in calculus courses. |
| 2 | Laplace transforms | In engineering and differential equations, partial fractions are essential for inverting Laplace transforms and solving linear systems. |
| 3 | Algebraic simplification | Sometimes breaking a complex expression into simpler parts makes other operations (like finding limits or series expansions) more manageable. |
The prerequisite for using partial fractions is that you can factor the denominator. If the denominator can't be factored over the real numbers, you may need to use irreducible quadratic factors, which we'll cover below.
Case 1: Distinct Linear Factors
This is the simplest and most common case. When the denominator factors into distinct linear terms (no repeated roots), each factor gets its own constant numerator.
Decompose (5x + 3)/((x + 1)(x − 2)).
Set up the form: (5x + 3)/((x + 1)(x − 2)) = A/(x + 1) + B/(x − 2).
Multiply both sides by (x + 1)(x − 2): 5x + 3 = A(x − 2) + B(x + 1).
Now use the cover-up method (plugging in roots): Set x = 2: 5(2) + 3 = A(0) + B(3), so 13 = 3B, giving B = 13/3.
Set x = −1: 5(−1) + 3 = A(−3) + B(0), so −2 = −3A, giving A = 2/3.
(5x + 3)/((x + 1)(x − 2)) = (2/3)/(x + 1) + (13/3)/(x − 2)The cover-up method works perfectly for distinct linear factors because plugging in each root eliminates all other terms, isolating one constant at a time.
Case 2: Repeated Linear Factors
When a linear factor appears more than once (e.g., (x − 1)² or (x + 3)³), you need a separate term for each power up to the multiplicity.
Decompose (2x + 1)/((x − 1)²(x + 2)).
Set up the form: A/(x − 1) + B/(x − 1)² + C/(x + 2). Note the two terms for (x − 1): one with denominator (x − 1) and one with (x − 1)².
Multiply both sides by (x − 1)²(x + 2): 2x + 1 = A(x − 1)(x + 2) + B(x + 2) + C(x − 1)².
Set x = 1: 3 = B(3), so B = 1.
Set x = −2: −3 = C(9), so C = −1/3.
For A, set x = 0: 1 = A(−1)(2) + 1(2) + (−1/3)(1), giving 1 = −2A + 2 − 1/3, so −2A = 1 − 2 + 1/3 = −2/3, hence A = 1/3.
A = 1/3, B = 1, C = −1/3(x − r) has multiplicity n, include n terms with denominators (x − r), (x − r)², ..., (x − r)ⁿ, each with its own constant numerator.Case 3: Irreducible Quadratic Factors
When the denominator contains a quadratic that can't be factored over the reals (discriminant < 0), the numerator of that term must be linear (Ax + B), not just a constant.
Decompose (x² + 3x + 5)/((x + 1)(x² + 1)). The factor x² + 1 is irreducible (discriminant = 0 − 4 = −4 < 0).
Set up the form: A/(x + 1) + (Bx + C)/(x² + 1).
Multiply both sides by (x + 1)(x² + 1): x² + 3x + 5 = A(x² + 1) + (Bx + C)(x + 1).
Set x = −1: 1 − 3 + 5 = A(2), so A = 3/2.
Expand the right side: (3/2)(x² + 1) + (Bx + C)(x + 1) = (3/2)x² + 3/2 + Bx² + Bx + Cx + C. Combine: (3/2 + B)x² + (B + C)x + (3/2 + C).
Match coefficients with the left side x² + 3x + 5: coefficient of x²: 3/2 + B = 1, so B = −1/2; coefficient of x: B + C = 3, so C = 7/2.
(3/2)/(x + 1) + (−x/2 + 7/2)/(x² + 1)The term with the irreducible quadratic integrates using a combination of natural log and arctangent.
Integrating After Decomposition
Once you've decomposed a rational expression, integration becomes straightforward. Here are the integration rules for each type of partial fraction term.
| Term Type | Integration Rule |
|---|---|
| Linear term | ∫ A/(x − r) dx = A ln|x − r| + C |
| Repeated linear term | ∫ A/(x − r)ⁿ dx = A · (x − r)^(1−n)/(1 − n) + C for n ≥ 2 |
| Irreducible quadratic | ∫ (Ax + B)/(x² + bx + c) dx splits into two integrals: one that gives a logarithm (match the derivative of the denominator) and one that gives an arctangent (complete the square). |
Evaluate ∫ (5x + 3)/((x + 1)(x − 2)) dx.
From our earlier decomposition: = ∫ (2/3)/(x + 1) dx + ∫ (13/3)/(x − 2) dx
= (2/3) ln|x + 1| + (13/3) ln|x − 2| + CWhat would have been a difficult integral became two basic logarithms — that's the power of partial fractions. If you're struggling with the decomposition step, scan the problem with Solver AI to see every step worked out clearly.