Partial fraction decomposition breaks a complicated rational function into a sum of simpler fractions that you can integrate one at a time. The technique applies whenever the integrand is a ratio of polynomials and the degree of the numerator is less than the degree of the denominator (if it is not, do polynomial long division first). Each factor of the denominator contributes a term to the decomposition.
Difficulty: Intermediate | Prerequisites: polynomial long division, factoring quadratics, basic logarithmic integration, inverse trig integrals.
Partial fractions is one of the core integration techniques taught alongside u-substitution, integration by parts, and trig substitution. It handles the case where you need to integrate a rational function (a polynomial divided by a polynomial). In the broader course arc, it typically comes right after or alongside trig substitution, and it rounds out the toolkit you need for virtually any integral you will encounter in the course. The technique rests on the algebraic fact that every rational function with real coefficients can be decomposed into a sum of simpler pieces.
Partial fraction decomposition
The process of expressing a rational function P(x)/Q(x), where deg(P) < deg(Q), as a sum of fractions whose denominators are the factors of Q(x). Think of it as: "breaking one hard fraction into several easy ones."
Proper rational function
A rational function where the degree of the numerator is strictly less than the degree of the denominator. In simple terms, this is the condition that must hold before you can decompose. If it does not hold, perform long division first.
Improper rational function
A rational function where the degree of the numerator is greater than or equal to the degree of the denominator. In simple terms, you need to do polynomial long division to pull out a polynomial part before decomposing the remainder.
Distinct linear factors
When the denominator factors into terms like (x − a)(x − b) with no repeats. Each factor contributes a term A/(x − a). Think of it as: the simplest case.
Repeated linear factors
When a linear factor appears more than once, e.g. x²(x − 2). The factor x² contributes both A/x and B/x², not just one term. In simple terms, each power of the repeated factor gets its own constant.
Irreducible quadratic factor
A quadratic factor like x² + x + 1 that cannot be factored further over the reals (its discriminant is negative). It contributes a term (Bx + C)/(x² + x + 1), with a linear numerator rather than a constant.
When the denominator factors into distinct linear terms, each factor gets a constant numerator.
Example (Q12): Decompose 18/(x² + 3x − 18).
Factor the denominator: x² + 3x − 18 = (x − 3)(x + 6). Write 18/((x − 3)(x + 6)) = A/(x − 3) + B/(x + 6). Multiply through: 18 = A(x + 6) + B(x − 3). Setting x = 3 gives A = 2; setting x = −6 gives B = −2. Result: 2/(x − 3) − 2/(x + 6).
Example (Q13): Find the term with denominator (x + 5) in the decomposition of (4x + 17)/(x² + 7x + 10).
Factor: (x + 5)(x + 2). Write A/(x + 5) + B/(x + 2). Equating coefficients: A + B = 4, 2A + 5B = 17, giving A = 1, B = 3. The term is 1/(x + 5).
When a factor appears more than once, you need a separate term for each power.
Example (Q21): Decompose (5x − 1)/(x²(x − 2)).
The factor x appears twice, so write A/x + B/x² + C/(x − 2). Multiply through: 5x − 1 = Ax(x − 2) + B(x − 2) + Cx². Setting x = 0 gives B = 1/2. Setting x = 2 gives C = 9/4. Comparing x² coefficients: A + C = 0, so A = −9/4. Result: −9/(4x) + 1/(2x²) + 9/(4(x − 2)).
When the denominator has a quadratic factor that does not factor over the reals, use a linear numerator (Bx + C).
Example (Q14): Decompose 12x/((x − 1)(x² + x + 1)).
Write A/(x − 1) + (Bx + C)/(x² + x + 1). Multiply through and equate coefficients: A + B = 0, A + C − B = 12, A − C = 0. Solving: A = 4, C = 4, B = −4. Result: 4/(x − 1) + (4 − 4x)/(x² + x + 1).
Once you have the decomposition, integrate each term individually. The key integrals that arise are:
∫ A/(x − a) dx = A ln|x − a| + C
∫ A/(x − a)² dx = −A/(x − a) + C
For irreducible quadratics, complete the square and use arctan.
When the degree of the numerator is not less than the degree of the denominator, divide first.
Example (Q15): Determine ∫ y/(y − 3) dy.
The degrees are equal, so divide: y/(y − 3) = 1 + 3/(y − 3). Now integrate: y + 3 ln|y − 3| + C.
Example (Q19): Determine ∫ (4x² − 5x + 3)/(x² − x − 2) dx.
Divide: the quotient is 4 with remainder −x + 11. Factor the denominator: (x + 1)(x − 2). Decompose (x − 11)/((x + 1)(x − 2)): gives 4/(x + 1) − 3/(x − 2). So the integral becomes 4x − 4 ln|x + 1| + 3 ln|x − 2| − (combine logs) = 4x − ln|(x + 1)⁴/(x − 2)³| + C.
Example (Q17): Evaluate ∫₃⁵ 2/((x − 2)(6 − x)) dx.
Decompose: A = 1/2, B = 1/2. Integrate: (1/2)[ln(x − 2) − ln(6 − x)] from 3 to 5. Evaluate: (1/2)(ln 3 − ln 1 − ln 1/3 + ln ... ) = (1/2) ln 9.
Example (Q20): Evaluate ∫₀^(ln 2) 3/(eˣ + 4) dx.
Substitute u = eˣ + 4, so du = (u − 4) dx. Limits: u = 5 to u = 6. The integral becomes ∫₅⁶ 3/(u(u − 4)) du. Decompose using partial fractions: 1/(u(u − 4)) = (1/4)(1/(u − 4) − 1/u). Integrate: (3/4)[ln|u − 4| − ln|u|]₅⁶ = (3/4)(ln(1/3) − ln(1/5)) = (3/4) ln(5/3).
Example (Q18): Find y satisfying dy/dx = 6/((x − 4)(9 − x)), y(5) = 0.
Decompose: both A and B equal 6/5. Integrate: y = (6/5)(ln|x − 4| − ln|9 − x|) + C = (6/5) ln|(x − 4)/(9 − x)| + C. Apply the initial condition y(5) = 0: C = (6/5) ln 4. Final answer: y = (6/5)(ln|(x − 4)/(9 − x)| + ln 4).
Denominator type | Form of decomposition |
|---|---|
Distinct linear: (x − a)(x − b) | A/(x − a) + B/(x − b) |
Repeated linear: (x − a)² | A/(x − a) + B/(x − a)² |
Repeated linear: (x − a)³ | A/(x − a) + B/(x − a)² + C/(x − a)³ |
Irreducible quadratic: (x² + bx + c) | (Ax + B)/(x² + bx + c) |
Mixed: (x − a)(x² + bx + c) | A/(x − a) + (Bx + C)/(x² + bx + c) |
Key integration results:
∫ 1/(x − a) dx = ln|x − a| + C
∫ 1/(x − a)² dx = −1/(x − a) + C
∫ 1/(x² + a²) dx = (1/a) arctan(x/a) + C
Students often forget to check whether the rational function is proper before decomposing. If deg(numerator) ≥ deg(denominator), you must do long division first. Skipping this step produces a decomposition that does not exist.
When a linear factor is repeated (e.g. x² in the denominator), students sometimes write only A/x² and miss the A/x term. Every power up to and including the highest gets its own fraction.
For irreducible quadratic factors, students often put a constant in the numerator instead of a linear expression. The numerator must be (Bx + C), not just B.
A frequent algebra error is plugging in values of x to find coefficients but then forgetting to check by expanding. Always verify your A, B, C values by multiplying back through and confirming the original numerator is recovered.
⚠️ Partial fractions is tested on virtually every integral calculus exam. Expect a mix of decomposition-only problems ("rewrite using partial fractions") and full integration problems.
⚠️ Problems that require long division before decomposition (like Q15 and Q19) are common. If you see degree(numerator) ≥ degree(denominator), divide first.
⚠️ The combination of u-substitution and partial fractions (like Q20, where u = eˣ + 4 leads to a partial fractions integral) is a favourite exam construction.
⚠️ Initial-value problems that use partial fractions for the integration step (Q18) connect this material to differential equations and are commonly tested.
True or False: You can apply partial fraction decomposition to (x³ + 1)/(x² − 1) directly without any preliminary step. (False: the degree of the numerator is greater, so you must do long division first.)
Fill in the blank: The partial fraction form of A/((x − 1)²(x + 3)) is ______. (A/(x − 1) + B/(x − 1)² + C/(x + 3))
True or False: For an irreducible quadratic factor x² + 4 in the denominator, the partial fraction numerator should be a constant A. (False: it should be Ax + B.)
Fill in the blank: ∫ 3/(x − 5) dx = ______. (3 ln|x − 5| + C)
True or False: To find coefficients in a partial fraction decomposition, you can substitute convenient values of x (such as the roots of each factor) into the equation. (True.)
Q: Decompose 18/(x² + 3x − 18) into partial fractions.
A: Factor as (x − 3)(x + 6). The decomposition is 2/(x − 3) − 2/(x + 6).
Q: Decompose 12x/((x − 1)(x² + x + 1)) into partial fractions.
A: 4/(x − 1) + (4 − 4x)/(x² + x + 1). The quadratic factor is irreducible, so its numerator is linear.
Q: Evaluate ∫ y/(y − 3) dy.
A: Divide first: y/(y − 3) = 1 + 3/(y − 3). Integrate: y + 3 ln|y − 3| + C.
Q: Find the indefinite integral ∫ (x + 11)/((x + 3)(x − 5)) dx.
A: Decompose: −1/(x + 3) + 2/(x − 5). Integrate: −ln|x + 3| + 2 ln|x − 5| + C = ln((x − 5)²/|x + 3|) + C.
Q: Evaluate ∫₃⁵ 2/((x − 2)(6 − x)) dx.
A: Decompose with A = B = 1/2. Integrate: (1/2) ln|(x − 2)/(6 − x)| from 3 to 5 = (1/2) ln 9.
Q: Decompose (5x − 1)/(x²(x − 2)) into partial fractions.
A: −9/(4x) + 1/(2x²) + 9/(4(x − 2)). Note the repeated factor x² requires both an x and an x² term.
Partial fractions complements trigonometric substitution: trig sub handles roots of quadratics, while partial fractions handles rational functions. Together with u-substitution and integration by parts, these four techniques cover nearly every integral in the course. Partial fractions also reappears in differential equations (separable ODEs often produce rational integrands) and in Laplace transforms, where inverse transforms rely heavily on partial fraction decomposition.
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