Partial Fractions: Repeated & Quadratics — Partial Fractions with Repeated and Irreducible Quadratics - Calculus 2

SECTION 1

Historical Context & Motivation

The technique of decomposing rational functions into sums of simpler fractions has roots stretching back to the earliest days of the calculus. When Leibniz and the Bernoulli brothers tackled integrals of rational functions in the late seventeenth century, they recognized that a fraction whose denominator factors into linear and quadratic pieces can be split into summands that are individually amenable to known antiderivative rules—logarithms and arctangent forms. The method of partial fractions thus arose not as an algebraic curiosity but as a practical necessity for evaluating integrals that appear throughout physics, engineering, and probability.

Simple partial fraction decompositions—those involving only distinct linear factors—are relatively straightforward. The real challenge emerges when the denominator contains repeated linear factors such as (x − a)³ or irreducible quadratic factors such as x² + 1 that cannot be factored further over the reals. These cases require additional terms in the decomposition and more sophisticated algebraic techniques to determine the unknown coefficients.

1702 • Bernoulli–Leibniz Correspondence

Johann Bernoulli and Leibniz debated whether every rational function could be integrated using partial fractions. Bernoulli initially believed that irreducible quadratic factors could always be avoided, but Leibniz correctly argued that real quadratic factors are sometimes irreducible.

1748 • Euler's Systematic Treatment

Leonhard Euler provided a complete and rigorous framework for partial fraction decomposition in his Introductio in analysin infinitorum, including the treatment of repeated factors and quadratic terms, establishing the technique as a standard tool of analysis.

1821 • Cauchy's Algebraic Foundations

Augustin-Louis Cauchy placed partial fractions on firm algebraic footing, connecting the technique to the Fundamental Theorem of Algebra and proving that every polynomial over the reals factors completely into linear and irreducible quadratic factors.

Modern Era • Integration and CAS

Partial fraction decomposition is now a core algorithm in computer algebra systems (Maple, Mathematica, SymPy) and remains an essential hand-calculation technique for students of calculus, differential equations, and Laplace transforms.

The central question this lesson addresses is: given a proper rational function P(x)/Q(x) whose denominator contains repeated linear factors, irreducible quadratic factors, or both, how do we systematically set up and solve the partial fraction decomposition so that we can integrate each piece individually?

SECTION 2

Core Principles & Definitions

Before diving into the mechanics, we need several foundational ideas. A rational function P(x)/Q(x) is called proper if deg(P) < deg(Q); if not, we first perform polynomial long division to extract a polynomial part plus a proper remainder. The Fundamental Theorem of Algebra guarantees that over the reals, Q(x) factors completely into products of linear factors (x − a) and irreducible quadratic factors (x² + bx + c) where b² − 4c < 0. An irreducible quadratic is one whose discriminant is negative, meaning it has no real roots and cannot be factored into real linear pieces.

Proper Fraction Requirement

Always ensure deg(P) < deg(Q) before decomposing. If not, perform polynomial long division first. The remainder fraction is the proper part to decompose.

Repeated Linear Factors

If (x − a)ⁿ divides Q(x), include n separate terms in the decomposition: A₁/(x − a) + A₂/(x − a)² + ⋯ + Aₙ/(x − a)ⁿ, with constant numerators.

Irreducible Quadratic Factors

For each irreducible quadratic factor (x² + bx + c), include a term (Ax + B)/(x² + bx + c). The numerator must be linear, not merely constant.

Repeated Irreducible Quadratics

If (x² + bx + c)ⁿ divides Q(x), include n terms with increasing powers in the denominator, each having a linear numerator: (A₁x + B₁)/(x² + bx + c) + ⋯ + (Aₙx + Bₙ)/(x² + bx + c)ⁿ.

Coefficient Determination

After setting up the decomposition, multiply both sides by Q(x) and solve for unknowns using strategic substitution (plugging in roots) and/or equating coefficients of like powers of x.

✦

KEY TAKEAWAY

Think of partial fraction decomposition as reverse-engineering a smoothie: the blended rational function P(x)/Q(x) is the smoothie, and your goal is to identify each individual ingredient (each simple fraction). Repeated factors are like having multiple scoops of the same fruit—you need a separate accounting for each scoop. An irreducible quadratic is like a two-component ingredient (say, peanut butter—protein and fat)—you must allow a two-parameter numerator (Ax + B) to capture both 'dimensions' of that factor.

SECTION 3

Visual Explanation — Decomposition Structure

The following diagram illustrates the decision tree for setting up a partial fraction decomposition. Starting from the factored denominator, each factor type leads to a specific template of terms in the expansion. Understanding this flowchart eliminates the most common source of errors: writing the wrong form for the decomposition.

The decision tree shows how each factor type in Q(x) dictates the form of terms in the partial fraction expansion. Linear factors produce constant numerators, while quadratic factors require linear numerators. Repeated factors of either type generate a cascade of terms up to the multiplicity.

Notice the critical structural rule visible in the diagram: the total number of unknown constants across all terms always equals deg(Q). This guarantees that multiplying through by Q(x) and equating coefficients yields a system with exactly as many equations as unknowns, so the decomposition is always uniquely determined.

SECTION 4

Mathematical Framework

We now formalize the decomposition templates and the integration formulas that result from each term type. Understanding these templates is essential because once the decomposition is established, each piece integrates via a standard formula.

Decomposition for Repeated Linear Factors

REPEATED LINEAR DECOMPOSITION

P(x) / [(x − a)ⁿ · R(x)] = A₁/(x − a) + A₂/(x − a)² + ⋯ + Aₙ/(x − a)ⁿ + [terms from R(x)]

Each Aₖ is a constant. The term Aₖ/(x − a)ᵏ integrates to Aₖ · (x − a)¹⁻ᵏ / (1 − k) for k ≥ 2, and to Aₖ ln|x − a| for k = 1.

Decomposition for Irreducible Quadratic Factors

IRREDUCIBLE QUADRATIC DECOMPOSITION

P(x) / [(x² + bx + c) · R(x)] = (Ax + B)/(x² + bx + c) + [terms from R(x)]

Here b² − 4c < 0. To integrate (Ax + B)/(x² + bx + c), complete the square in the denominator and split the numerator into a part proportional to the derivative of the denominator (yielding a logarithm) and a constant part (yielding an arctangent).

Integration of a Quadratic Term

QUADRATIC INTEGRATION FORMULA

∫ (Ax + B) / (x² + bx + c) dx = (A/2) ln(x² + bx + c) + [(2B − Ab) / √(4c − b²)] arctan[(2x + b) / √(4c − b²)] + C

This formula follows from writing x² + bx + c = (x + b/2)² + (4c − b²)/4, then splitting Ax + B = (A/2)(2x + b) + (B − Ab/2). The first piece is a du/u integral and the second is an arctan integral after substitution.

REPEATED QUADRATIC INTEGRATION (REDUCTION)

∫ (Ax + B) / (x² + bx + c)ⁿ dx (n ≥ 2)

For the du/u-type piece, the integral is (A/2) · (x² + bx + c)¹⁻ⁿ / (1 − n). The arctangent piece requires a trigonometric substitution or a reduction formula. In practice these appear less frequently; the key point is that the decomposition form with linear numerators is correct and guaranteed to lead to elementary antiderivatives.

💡

Coefficient Strategies

After multiplying through by Q(x), you can find the unknowns by: (1) strategic substitution—plug in x = a for each real root a of Q(x) to isolate certain constants; (2) equating coefficients—expand both sides and match coefficients of xⁿ, xⁿ⁻¹, … to get a linear system. For irreducible quadratics, equating coefficients is usually necessary since there are no real roots to substitute.

SECTION 5

Detailed Classification of Factor Types

The table below summarizes every factor type, the corresponding decomposition template, and the resulting antiderivative form. This reference is the single most useful resource for setting up partial fraction problems correctly.

Factor Type	Decomposition Template	Antiderivative Form
Distinct linear (x − a)	A / (x − a)	A ln\|x − a\|
Repeated linear (x − a)ⁿ	A₁/(x−a) + A₂/(x−a)² + ⋯ + Aₙ/(x−a)ⁿ	Logs for k=1; power rule Aₖ(x−a)¹⁻ᵏ/(1−k) for k≥2
Distinct irreducible quadratic (x²+bx+c)	(Ax+B) / (x²+bx+c)	Logarithm + arctangent (after completing the square)
Repeated irreducible quadratic (x²+bx+c)ⁿ	Σₖ (Aₖx+Bₖ)/(x²+bx+c)ᵏ for k=1…n	Power rule for du/u part; reduction/trig-sub for arctan part

Complete classification of partial fraction factor types and their integration outcomes

This diagram dissects the decomposition of a rational function with a repeated linear factor (x − 1)² and an irreducible quadratic factor (x² + 4). Note the four unknowns match the degree of Q(x), and each term's integration formula is shown at the bottom.

The diagram above highlights two important points. First, the repeated factor (x − 1)² generates two terms with constant numerators—one for each power from 1 to 2. Second, the irreducible quadratic factor x² + 4 generates a single term, but with a linear numerator Cx + D rather than just a constant. Forgetting the linear numerator for quadratic factors is one of the most common mistakes students make.

SECTION 6

Worked Example — Complete Decomposition & Integration

Let us work through a complete example that involves both a repeated linear factor and an irreducible quadratic factor, culminating in the evaluation of the integral.

Evaluate ∫ (4x² + 3x − 1) / [(x − 2)²(x² + 1)] dx

1
Step 1 — Verify Proper Fraction and Set Up Decomposition

The numerator has degree 2 and the denominator has degree 4, so the fraction is proper. The denominator factors as (x − 2)² · (x² + 1). The factor (x − 2) is linear and repeated with multiplicity 2, and (x² + 1) is an irreducible quadratic (discriminant = 0² − 4(1)(1) = −4 < 0). We write: (4x² + 3x − 1) / [(x − 2)²(x² + 1)] = A/(x − 2) + B/(x − 2)² + (Cx + D)/(x² + 1)

Four unknowns: A, B, C, D (matches deg(Q) = 4 ✓)
2
Step 2 — Multiply Both Sides by (x − 2)²(x² + 1)

Clearing denominators gives: 4x² + 3x − 1 = A(x − 2)(x² + 1) + B(x² + 1) + (Cx + D)(x − 2)²
3
Step 3 — Strategic Substitution (x = 2)

Setting x = 2 eliminates terms containing (x − 2): 4(4) + 3(2) − 1 = A(0)(5) + B(5) + (2C + D)(0) 16 + 6 − 1 = 5B 21 = 5B

B = 21/5
4
Step 4 — Equate Coefficients to Find A, C, D

We expand the right side. A(x − 2)(x² + 1) = A(x³ − 2x² + x − 2). B(x² + 1) = Bx² + B. (Cx + D)(x − 2)² = (Cx + D)(x² − 4x + 4) = Cx³ − 4Cx² + 4Cx + Dx² − 4Dx + 4D. Collecting by powers of x: x³: A + C = 0 x²: −2A + B − 4C + D = 4 x¹: A + 4C − 4D = 3 x⁰: −2A + B + 4D = −1 From x³: C = −A. Substituting B = 21/5 into x⁰: −2A + 21/5 + 4D = −1, so −2A + 4D = −26/5. From x¹: A − 4A − 4D = 3, so −3A − 4D = 3. Adding the last two: −5A = 3 − 26/5 = −11/5, so A = 11/25. Then C = −11/25, and from −3(11/25) − 4D = 3: −33/25 − 4D = 3, so 4D = −33/25 − 75/25 = −108/25, giving D = −27/25.

A = 11/25, B = 21/5, C = −11/25, D = −27/25
5
Step 5 — Integrate Each Term

∫ A/(x − 2) dx = (11/25) ln|x − 2| ∫ B/(x − 2)² dx = (21/5) · (−1)/(x − 2) = −21/[5(x − 2)] ∫ (Cx + D)/(x² + 1) dx: Split as C∫ x/(x² + 1) dx + D∫ 1/(x² + 1) dx = (−11/25) · (1/2) ln(x² + 1) + (−27/25) arctan(x) = −(11/50) ln(x² + 1) − (27/25) arctan(x)

∫ = (11/25) ln|x−2| − 21/[5(x−2)] − (11/50) ln(x²+1) − (27/25) arctan(x) + C

SECTION 7

Comparison of Coefficient-Finding Strategies

When solving for the unknown coefficients in a partial fraction decomposition, students have several methods at their disposal. Each has distinct advantages and limitations depending on the factor structure of the denominator.

Method	Strengths	Limitations
Strategic Substitution	Instantly determines coefficients associated with linear factors. Substitute x = a to isolate the constant for the highest-power term of (x − a).	Cannot directly isolate coefficients for irreducible quadratic terms (no real roots to substitute). Only isolates the coefficient of the highest-power repeated term directly.
Equating Coefficients	Works universally for all factor types. Produces a complete linear system that is always solvable.	Requires expanding and collecting terms, which is algebraically intensive for high-degree denominators. Error-prone in hand calculations.
Hybrid (Substitution + Coefficients)	Use substitution first to find as many constants as possible, then equate coefficients for the rest. Minimizes algebra.	Requires judgment about when to switch methods. Students may lose track of which unknowns are already determined and which remain to be found, especially when the denominator has many factors.
Heaviside Cover-Up	Extremely fast for distinct linear factors: mentally 'cover' (x − a) and evaluate at x = a.	Does not apply to repeated factors (except the highest-power term) or to quadratic factors at all.

Comparison of methods for determining unknown coefficients

🎯

BEST PRACTICE

In practice, the hybrid approach is almost always optimal. Think of it like solving a jigsaw puzzle: strategic substitution gives you the edge pieces quickly, and equating coefficients fills in the interior. Start by substituting every real root to harvest all the easy constants, then use just enough coefficient equations to finish the job. This strategy minimizes both computation and the chance of algebraic error.

SECTION 8

Connections to Advanced Topics

Partial fraction decomposition is not merely a computational trick for Calculus 2—it is a fundamental algebraic technique with deep connections to several advanced mathematical and engineering topics. Understanding where partial fractions lead motivates why mastering the technique now pays significant dividends later.

Calculus 2 Context	Advanced Application
Decomposing P(x)/Q(x) to evaluate ∫ P(x)/Q(x) dx	Inverse Laplace transforms: decompose F(s) = P(s)/Q(s) to find f(t) term by term in differential equations and control theory
Irreducible quadratic terms producing arctan antiderivatives	Residue calculus in complex analysis: irreducible quadratics correspond to conjugate pairs of complex poles, and partial fractions over ℂ use only linear terms
Repeated factors generating power-rule antiderivatives	Z-transform inversion in discrete signals: repeated poles in H(z) require analogous partial fraction expansions for time-domain recovery
Counting unknowns = deg(Q) ensuring unique decomposition	Algebraic geometry: partial fractions relate to the Mittag-Leffler theorem, which generalizes the idea to meromorphic functions on Riemann surfaces

From Calculus 2 partial fractions to advanced mathematical applications

Perhaps the most immediate connection is to Laplace transforms in a differential equations course. When solving a linear ODE with constant coefficients via the Laplace transform, you obtain an algebraic expression Y(s) = P(s)/Q(s) in the s-domain. To invert back to the time domain, you decompose Y(s) into partial fractions—exactly the same procedure studied here—and then read off each term's inverse transform from a standard table. Repeated factors produce terms involving tⁿeᵃᵗ, and irreducible quadratic factors produce damped sinusoidal terms eᵅᵗ sin(βt) or eᵅᵗ cos(βt), which are the fundamental building blocks of solutions to engineering dynamical systems.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why the partial fraction decomposition of P(x)/[(x² + 1)²] requires numerators of the form (Ax + B)/(x² + 1) + (Cx + D)/(x² + 1)² rather than simple constants A/(x² + 1) + B/(x² + 1)². What would go wrong if you used only constants in the numerators?

Show answer

The decomposition must have enough free parameters to match the original numerator, which can have degree up to 3 (one less than deg(Q) = 4). Using only constants A and B in the numerators gives just 2 parameters, but the system requires 4 to match all coefficients. More fundamentally, the fraction (Ax + B)/(x² + 1) has a linear numerator because the denominator is degree 2—we need a numerator of degree one less than each irreducible factor's degree. If we used only constants, we could never reproduce a numerator with an x-term, making the decomposition impossible for most choices of P(x). The general rule is: each irreducible quadratic factor of multiplicity k contributes terms with linear numerators (Aᵢx + Bᵢ) for i = 1, …, k.

Question 2: BASIC CALCULATION

Find the partial fraction decomposition of (5x + 3) / [(x + 1)²(x − 2)].

Show answer

Set up: (5x + 3)/[(x+1)²(x−2)] = A/(x+1) + B/(x+1)² + C/(x−2). Multiply through: 5x + 3 = A(x+1)(x−2) + B(x−2) + C(x+1)². Substituting x = −1: 5(−1)+3 = B(−3), so −2 = −3B, giving B = 2/3. Substituting x = 2: 13 = C(9), so C = 13/9. Equating x² coefficients: 0 = A + C, so A = −13/9. The decomposition is (−13/9)/(x+1) + (2/3)/(x+1)² + (13/9)/(x−2). You can verify by recombining over the common denominator.

Question 3: INTERMEDIATE

Evaluate ∫ (x² + 2) / [x(x² + 4)] dx.

Show answer

Decompose: (x² + 2)/[x(x² + 4)] = A/x + (Bx + C)/(x² + 4). Multiplying through: x² + 2 = A(x² + 4) + (Bx + C)x = (A + B)x² + Cx + 4A. Equating coefficients: x²: A + B = 1; x¹: C = 0; x⁰: 4A = 2, so A = 1/2. Then B = 1/2 and C = 0. The integral becomes ∫ (1/2)/x dx + ∫ (x/2)/(x² + 4) dx = (1/2) ln|x| + (1/4) ln(x² + 4) + K. The second integral was evaluated via the substitution u = x² + 4, du = 2x dx. Notice that C = 0 means no arctangent term appears in this particular problem.

Question 4: APPLIED

In a circuit with a second-order transfer function, the Laplace-domain voltage is V(s) = (3s + 5) / [(s + 1)(s² + 2s + 5)]. Decompose V(s) into partial fractions to prepare for inverse Laplace transformation. (Note: s² + 2s + 5 is irreducible since its discriminant is 4 − 20 = −16 < 0.)

Show answer

Set up: (3s+5)/[(s+1)(s²+2s+5)] = A/(s+1) + (Bs+C)/(s²+2s+5). Multiply through: 3s + 5 = A(s²+2s+5) + (Bs+C)(s+1). Substituting s = −1: 3(−1)+5 = A(1−2+5) = 4A, so 2 = 4A, giving A = 1/2. Equating s² coefficients: 0 = A + B = 1/2 + B, so B = −1/2. Equating s⁰ coefficients: 5 = 5A + C = 5/2 + C, so C = 5/2. The decomposition is (1/2)/(s+1) + (−s/2 + 5/2)/(s²+2s+5). For the inverse Laplace transform, one would complete the square: s²+2s+5 = (s+1)²+4, then rewrite the quadratic term as [−(s+1)/2 + 3]/[(s+1)²+4], separating into a damped cosine and a damped sine component.

Question 5: CRITICAL THINKING

Consider the integral ∫ 1 / [(x² + 1)²] dx. Set up the partial fraction decomposition (which has the form (Ax + B)/(x² + 1) + (Cx + D)/(x² + 1)²), and explain why this decomposition is trivial (i.e., A = B = 0, C = 0, D = 1). Then discuss: if partial fractions don't simplify this integral, what technique should be used instead? Evaluate the integral.

Show answer

The decomposition 1/(x²+1)² = (Ax+B)/(x²+1) + (Cx+D)/(x²+1)² leads to 1 = (Ax+B)(x²+1) + (Cx+D). Equating: x³: A = 0; x²: B = 0; x¹: C = 0 (using A=0); x⁰: B+D = 1, so D = 1. The decomposition simply recovers the original integrand—partial fractions provide no simplification when the denominator is a single irreducible factor raised to a power. Instead, use the trigonometric substitution x = tan θ, dx = sec²θ dθ. Then x²+1 = sec²θ, so 1/(x²+1)² = cos⁴θ/1 (wait—more carefully: 1/sec⁴θ = cos⁴θ). The integral becomes ∫ cos⁴θ · sec²θ dθ/ sec²θ... Let's be precise: ∫ dx/(x²+1)² = ∫ sec²θ dθ / sec⁴θ = ∫ cos²θ dθ = (1/2)(θ + sinθ cosθ) + C. Converting back: θ = arctan x, sinθ = x/√(x²+1), cosθ = 1/√(x²+1). So the result is (1/2)[arctan x + x/(x²+1)] + C. This problem illustrates that partial fractions are not a universal tool; when the denominator is already a single irreducible power, other techniques like trigonometric substitution or reduction formulas are needed.

SUMMARY

Summary & Key Concepts

Summary

Partial fraction decomposition transforms a proper rational function P(x)/Q(x) into a sum of simpler fractions that can each be integrated individually. When Q(x) contains repeated linear factors (x − a)ⁿ, the decomposition includes n terms A₁/(x − a) + A₂/(x − a)² + ⋯ + Aₙ/(x − a)ⁿ, each with a constant numerator. When Q(x) contains an irreducible quadratic factor (x² + bx + c) with b² − 4c < 0, the corresponding term has a linear numerator (Ax + B). Repeated irreducible quadratics combine both ideas: each power from 1 to n receives its own term with a linear numerator.

To find the unknown coefficients, use a hybrid strategy: apply strategic substitution at real roots first, then equate coefficients of like powers to determine the remaining unknowns. The total number of unknowns always equals deg(Q), guaranteeing a uniquely solvable system. Integration of the resulting terms yields logarithms from linear denominators, arctangent terms from irreducible quadratic denominators (after completing the square), and power-rule antiderivatives from repeated factors with k ≥ 2. This technique extends naturally into Laplace transforms, complex analysis, and signal processing.