Lesson 15: Partial Fractions and Rational Function Decomposition

Partial fraction decomposition runs a single complicated rational function backwards into a sum of simple ones. It looks like an algebraic party trick, but it exists for one reason: in calculus, sums of simple fractions (1/(x − a), 1/(x − a)^2, (Bx + C)/(x^2 + c)) are exactly the pieces we know how to integrate. This lesson builds the decomposition reliably for every denominator type.

The set-up: degree check and factor types

First, if deg(numerator) ≥ deg(denominator), divide: the quotient is a polynomial and the remainder is a proper fraction to decompose. Then factor the denominator completely over the reals. Each factor type contributes a standard term:

• Distinct linear factor (x − a): a term A/(x − a)
• Repeated linear factor (x − a)^k: terms A_1/(x − a) + A_2/(x − a)^2 + ... + A_k/(x − a)^k — one for every power
• Irreducible quadratic factor (x^2 + bx + c): a term (Bx + C)/(x^2 + bx + c) — with a linear numerator

Worked example (improper): (x^2 + 3x + 5)/(x + 1). Division: x^2 + 3x + 5 = (x + 1)(x + 2) + 3, so the fraction equals x + 2 + 3/(x + 1). Skipping the division is the most common structural error — the decomposition templates only fit proper fractions.

Distinct linear factors: the cover-up method

Worked example: decompose (5x − 4)/((x − 1)(x + 2)) = A/(x − 1) + B/(x + 2). Multiply through by (x − 1)(x + 2): 5x − 4 = A(x + 2) + B(x − 1). Substitute the roots of each factor ("cover up" the factor and evaluate): at x = 1: 5 − 4 = 1 = A(3), so A = 1/3; at x = −2: −10 − 4 = −14 = B(−3), so B = 14/3. Verify by recombining: x-coefficient A + B = 1/3 + 14/3 = 5 ✓; constant 2A − B = 2/3 − 14/3 = −4 ✓. The cover-up works because at x = 1 the B-term vanishes (x − 1 = 0) — choosing evaluation points that kill all but one unknown is the whole trick, and it extends to any distinct linear factors.

Repeated factors and irreducible quadratics

Repeated factor: (3x + 5)/(x − 1)^2 = A/(x − 1) + B/(x − 1)^2. Multiply through by (x − 1)^2: 3x + 5 = A(x − 1) + B. At x = 1: 8 = B. Then A: compare x-coefficients (A = 3) or evaluate at another x (x = 0: 5 = −A + 8 → A = 3 ✓). Both terms must be present — omitting A/(x − 1) makes the system unsolvable, the structural error this case is designed to catch. Irreducible quadratic: (2x + 1)/(x(x^2 + 1)) = A/x + (Bx + C)/(x^2 + 1). Clearing denominators: 2x + 1 = A(x^2 + 1) + (Bx + C)x = (A + B)x^2 + Cx + A. Match coefficients: A = 1 (constant), C = 2 (x-term), A + B = 0 so B = −1. Result: 1/x + (2 − x)/(x^2 + 1). For quadratics the cover-up is unavailable (no real root), so coefficient comparison — or substituting convenient x values — does the work.

Why this matters: the calculus connection

Each template piece has a known integral: ∫ dx/(x − a) = ln|x − a|, ∫ dx/(x − a)^2 = −1/(x − a), and (Bx + C)/(x^2 + c) splits into a logarithmic part and an arctangent part. Decomposing first is *the* method for integrating rational functions — a messy integrand like (5x − 4)/((x − 1)(x + 2)) becomes (1/3)/(x − 1) + (14/3)/(x + 2), two lookups instead of an impossible integral. The decomposition is also unique: a proper rational function has exactly one partial fraction form, which is why verification by recombining is a complete proof.

Common mistakes

• Forgetting to divide first. If the numerator's degree is not smaller, no template will fit — symptoms include a coefficient system that refuses to solve. Check degrees before anything else.
• Missing a term for a repeated factor. (x − 1)^2 needs both A/(x − 1) and B/(x − 1)^2. With only the squared term, matching coefficients is impossible — as it should be, since the form is wrong.
• Constant numerator over a quadratic. The numerator over an irreducible quadratic is Bx + C, not a constant B. Using a constant loses a degree of freedom and the system becomes inconsistent.
• Cover-up at non-roots. The cover-up substitution must use the root of the factor being isolated (x = 1 for (x − 1)); evaluating at x = 0 leaves both unknowns alive and forces simultaneous equations — slower and error-prone.
• Sign slips in cover-up arithmetic. At x = −2, the factor (x − 1) evaluates to −3, not 3: B = −14/(−3) = 14/3. Every cover-up step needs its signs written out.

Exam Experience

1. The correct partial fraction form for (7x − 1)/((x + 2)(x^2 + 3)) is

• (A) A/(x + 2) + B/(x^2 + 3)
• (B) A/(x + 2) + (Bx + C)/(x^2 + 3)
• (C) A/(x + 2) + B/(x^2 + 3) + C/(x + 2)
• (D) (Ax + B)/(x + 2) + C/(x^2 + 3)

Answer: (B). A linear factor takes a constant numerator; an irreducible quadratic takes a linear numerator Bx + C. (A) undercounts the unknowns (2 constants cannot match 3 coefficients); (D) puts the linear numerator over the wrong factor.

2. In (4x + 3)/((x − 1)(x + 3)) = A/(x − 1) + B/(x + 3), the value of A is

• (A) 7/4
• (B) 4/3
• (C) 7/3
• (D) 9/4

Answer: (A). Cover up (x − 1) and set x = 1: A = (4·1 + 3)/(1 + 3) = 7/4. (D) is the value of B (cover up (x + 3) at x = −3: (−12 + 3)/(−4) = 9/4) — a distractor for students who isolate the wrong factor.

3. Which fraction must be divided (long division) before setting up a decomposition?

• (A) (2x + 1)/(x^2 − 4)
• (B) (x^3 + 1)/(x^2 + 1)
• (C) (5)/( (x − 1)(x − 2) )
• (D) (x + 4)/(x^2 + x + 1)

Answer: (B). Numerator degree 3 ≥ denominator degree 2 — improper. The others are proper (numerator degree strictly smaller), so they go straight to templates.

Constructed response. Decompose (4x^2 − 7x − 12)/(x(x − 2)(x + 3)) completely. (a) Write the template form with constants A, B, C. (b) Use the cover-up method to find each constant, showing the substitution for each. (c) Verify your decomposition by recombining, or by evaluating both sides at x = 1.

Scoring guide. (a) 1 point: A/x + B/(x − 2) + C/(x + 3). (b) 3 points (1 each): clearing denominators gives 4x^2 − 7x − 12 = A(x − 2)(x + 3) + Bx(x + 3) + Cx(x − 2). At x = 0: −12 = A(−2)(3) = −6A, so A = 2. At x = 2: 16 − 14 − 12 = −10 = B(2)(5) = 10B, so B = −1. At x = −3: 36 + 21 − 12 = 45 = C(−3)(−5) = 15C, so C = 3. (c) 1 point: at x = 1 the original is (4 − 7 − 12)/(1 × (−1) × 4) = −15/−4 = 15/4; the decomposition gives 2/1 + (−1)/(−1) + 3/4 = 2 + 1 + 3/4 = 15/4 ✓.

Going deeper: a mixed denominator, fully worked

Worked example: decompose (x^2 + 2x + 3)/((x − 1)(x^2 + 4)) — one distinct linear factor, one irreducible quadratic. Template: A/(x − 1) + (Bx + C)/(x^2 + 4). Clear denominators:

x^2 + 2x + 3 = A(x^2 + 4) + (Bx + C)(x − 1) = (A + B)x^2 + (C − B)x + (4A − C)

Match coefficients: A + B = 1; C − B = 2; 4A − C = 3. From the first, B = 1 − A; from the third, C = 4A − 3. Substituting in the second: (4A − 3) − (1 − A) = 2 → 5A − 4 = 2 → A = 6/5, hence B = −1/5 and C = 9/5. Verify at x = 0: the original is 3/((−1)(4)) = −3/4; the decomposition gives (6/5)/(−1) + (9/5)/4 = −6/5 + 9/20 = −15/20 = −3/4 ✓. Three equations, three unknowns, solved by substitution — for mixed denominators, coefficient matching (or a hybrid: cover-up for the linear factor's A where the factor has a real root — here setting x = 1 in the cleared equation gives 1 + 2 + 3 = A(1 + 4) = 5A, so A = 6/5 immediately) is the reliable path.

And the destination, one last time: ∫ (5x − 4)/((x − 1)(x + 2)) dx = ∫ [(1/3)/(x − 1) + (14/3)/(x + 2)] dx = (1/3) ln|x − 1| + (14/3) ln|x + 2| + C. Decomposition converts an un-integrable fraction into a sum of standard forms. That conversion — not the algebra for its own sake — is why this technique survives in every calculus syllabus.

Workshop: one more decomposition, checked two ways

Worked example: (x + 2)/((x − 3)(x + 1)) = A/(x − 3) + B/(x + 1). Cover up: at x = 3, A = (3 + 2)/(3 + 1) = 5/4; at x = −1, B = (−1 + 2)/(−1 − 3) = 1/(−4) = −1/4. Verify by recombining: A(x + 1) + B(x − 3) = (5/4)(x + 1) − (1/4)(x − 3) = (5x + 5 − x + 3)/4 = (4x + 8)/4 = x + 2 ✓ — the numerator is recovered exactly, so the decomposition is proved, not merely plausible. Build the verification into your workflow permanently: recombination takes under a minute and converts "I think" into "I checked" — and the coefficient cross-check (A + B must equal the x-coefficient of the numerator, here 5/4 − 1/4 = 1 ✓) takes ten seconds.