Partial Fractions Decomposition

⚡ Key Takeaways

Partial Fractions Decomposition rewrites a single complicated rational function (a polynomial divided by a polynomial) as a sum of simpler fractions whose denominators are the factors of the original denominator. Key result: any proper rational function with a factorable denominator equals a sum of terms of the form $\frac{A}{(x-r)^k}$ and $\frac{Bx+C}{(x^2+bx+c)^k}$ . Applies in: integration, inverse Laplace transforms, infinite series, control theory, and algebra. Origin: developed by Johann Bernoulli and Leibniz in the early 1700s for integrating rational functions.

💡 The Core Insight

Adding fractions combines many pieces into one by finding a common denominator; partial fractions runs this process in reverse. If you can see the denominator's factors, you can "un-add" the fraction back into the simple parts it came from. Because a polynomial of degree $n$ factors into $n$ linear/quadratic building blocks, the original fraction is a disguised sum — and decomposition recovers the disguise.

DefinitionRational Function and Properness

A rational function is a quotient

1

\quad R(x) = \frac{P(x)}{Q(x)}, \qquad Q(x) \neq 0,

Derivation hint: $P$ and $Q$ are polynomials; the function is undefined at the roots of $Q$ . where $P(x)$ and $Q(x)$ are polynomials. The function is proper if $\deg P < \deg Q$ , and improper if $\deg P \geq \deg Q$ . Symbols: $\deg$ denotes the degree (highest power of $x$ ); $r$ denotes a root; $A, B, C$ denote unknown constants to be found.

DefinitionThe Four Standard Forms

Every proper rational function decomposes into a sum of terms drawn from these four types, depending on the factor in the denominator:

2

\quad \frac{A}{x - r} \quad\text{(distinct linear factor)}

Derivation hint: arises from a single, non-repeated root $r$ of $Q$ .

3

\quad \frac{A_1}{x-r} + \frac{A_2}{(x-r)^2} + \cdots + \frac{A_k}{(x-r)^k} \quad\text{(repeated linear factor)}

Derivation hint: a factor $(x-r)^k$ requires one term for each power from $1$ to $k$ .

4

\quad \frac{Bx + C}{x^2 + bx + c} \quad\text{(irreducible quadratic factor)}

Derivation hint: the numerator is linear (one degree below the quadratic) because $b^2 - 4c < 0$ prevents real factoring.

5

\quad \frac{B_1 x + C_1}{x^2+bx+c} + \cdots + \frac{B_m x + C_m}{(x^2+bx+c)^m} \quad\text{(repeated quadratic)}

Derivation hint: each power of the repeated quadratic gets its own linear-over-quadratic term.

TheoremExistence and Uniqueness of Decomposition

Let $R(x) = \frac{P(x)}{Q(x)}$ be a proper rational function with real coefficients, and let $Q$ be factored completely into distinct linear factors $(x - r_i)^{k_i}$ and distinct irreducible quadratic factors $(x^2 + b_j x + c_j)^{m_j}$ . Then there exist unique real constants such that $R(x)$ equals the sum of the corresponding standard-form terms [2]–[5]. If $R$ is improper, polynomial long division first yields

6

\quad \frac{P(x)}{Q(x)} = S(x) + \frac{P^*(x)}{Q(x)}, \qquad \deg P^* < \deg Q,

Derivation hint: divide $P$ by $Q$ ; $S$ is the quotient polynomial and $P^*$ the remainder, which is now proper. where $S(x)$ is a polynomial and the remaining proper fraction decomposes as above.

ProofSketch of Existence via Dimension Counting

We show why the standard forms always provide enough freedom to match any proper numerator. 1. Count the unknowns. Suppose $\deg Q = n$ . Each distinct linear factor contributes one constant; a $k$ -fold linear factor contributes $k$ constants; each $m$ -fold quadratic factor contributes $2m$ constants. Summing across all factors gives exactly $n$ unknown constants. 2. Count the equations. Clearing denominators produces a polynomial identity of degree at most $n-1$ (since $R$ is proper). A polynomial of degree $n-1$ has $n$ coefficients, so matching it produces exactly $n$ linear equations. 3. Match counts. We have $n$ unknowns and $n$ equations — a square linear system. The Fundamental Theorem of Algebra guarantees the factorization of $Q$ exists, and the resulting system can be shown to be nonsingular, so a unique solution exists. 4. Uniqueness. If two decompositions existed, their difference would be a proper rational function equal to zero, forcing all its constants to vanish — contradiction unless they were identical. $\blacksquare$

∎

mathematician · Johann Bernoulli | 1667–1748

Johann Bernoulli, a central figure of the early calculus era, systematically developed the integration of rational functions in his lectures (later compiled by L'Hôpital). He recognized that every rational function could be integrated in closed form by first splitting it into partial fractions and integrating each elementary piece — yielding logarithms, arctangents, and rational terms. This insight, shared with Leibniz, transformed integration from an art into an algorithm for the entire class of rational functions and cemented partial fractions as a foundational calculus tool.

🧠 Intuition

Think of a smoothie. The combined fraction $\frac{5x-4}{x^2 - x - 2}$ is the smoothie — blended, you cannot tell what went in. Partial fractions is the recipe card that tells you it was made from $\frac{2}{x+1}$ and $\frac{3}{x-2}$ . The denominator $x^2 - x - 2 = (x+1)(x-2)$ reveals the ingredient list (the factors); the numerators are the amounts. Because adding $\frac{2}{x+1} + \frac{3}{x-2}$ over a common denominator returns exactly the original, you know the recipe is correct. The whole technique is just "reverse blending": find the factors, then solve for how much of each pure fraction recombines to the given mix.

Figure

Reversing the addition of fractions: combining versus decomposing

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ExampleDistinct Linear Factors

Given: Decompose $\dfrac{5x - 4}{x^2 - x - 2}$ .

Step 1 — Factor the denominator.

7

\quad x^2 - x - 2 = (x+1)(x-2).

Derivation hint: find two numbers multiplying to $-2$ and adding to $-1$ : namely $+1$ and $-2$ .

Step 2 — Set up the form.

8

\quad \frac{5x-4}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}.

Derivation hint: one unknown constant per distinct linear factor.

Step 3 — Clear denominators.

9

\quad 5x - 4 = A(x-2) + B(x+1).

Derivation hint: multiply both sides by $(x+1)(x-2)$ .

Step 4 — Cover-up substitution. Set $x = 2$ : $\;6 = B(3) \Rightarrow B = 2$ . Set $x = -1$ : $\;-9 = A(-3) \Rightarrow A = 3$ .

Answer:

\boxed{\dfrac{5x-4}{x^2-x-2} = \dfrac{3}{x+1} + \dfrac{2}{x-2}}

Check: $\frac{3(x-2)+2(x+1)}{(x+1)(x-2)} = \frac{5x-4}{(x+1)(x-2)}$ . ✓

ExampleRepeated Linear and Quadratic Factors

Given: Decompose $\dfrac{2x^2 + 3}{(x-1)^2(x^2+1)}$ .

Step 1 — Set up the full form. A squared linear factor needs two terms; an irreducible quadratic needs a linear numerator:

10

\quad \frac{2x^2+3}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}.

Derivation hint: forms [3] and [4] combined; four unknowns match $\deg Q = 4$ .

Step 2 — Clear denominators.

11

\quad 2x^2+3 = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)^2.

Derivation hint: multiply through by $(x-1)^2(x^2+1)$ .

Step 3 — Use convenient values. Set $x=1$ : $\;5 = B(2) \Rightarrow B = \tfrac{5}{2}$ .

Step 4 — Equate coefficients. Expanding and collecting:

$x^3$ : $\;0 = A + C$
$x^2$ : $\;2 = -A + B + (-2C + D)$
$x^1$ : $\;0 = A - C + (C - 2D)$ , i.e. $0 = A - 2D$ ...

Solving with $B = \tfrac52$ gives $A = \tfrac12,\; C = -\tfrac12,\; D = \tfrac14$ .

Answer:

\boxed{\dfrac{2x^2+3}{(x-1)^2(x^2+1)} = \dfrac{1/2}{x-1} + \dfrac{5/2}{(x-1)^2} + \dfrac{-\tfrac12 x + \tfrac14}{x^2+1}}

⚠️ Always Check Properness and Irreducibility First

Two of the most frequent errors:

Forgetting to divide first. If $\deg P \geq \deg Q$ (improper), you must do polynomial long division before decomposing, as in [6]. Skipping this gives a system with no solution.
Putting a constant over a quadratic. An irreducible quadratic factor needs a linear numerator $Bx + C$ , not just a constant $B$ . Using only a constant loses a degree of freedom and the system fails.

CounterexampleA Wrong Setup That Cannot Work

Suppose a student writes

12

\quad \frac{x}{(x^2+1)^2} \overset{?}{=} \frac{A}{x^2+1} + \frac{B}{(x^2+1)^2}.

Derivation hint: this omits the linear numerators required by form [5]. This is wrong: a repeated irreducible quadratic requires linear numerators at every power. The correct form is

13

\quad \frac{x}{(x^2+1)^2} = \frac{A_1 x + B_1}{x^2+1} + \frac{A_2 x + B_2}{(x^2+1)^2}.

Derivation hint: solving gives $A_1=B_1=A_2=0,\; B_2=1$ — but the constant-only setup [12] cannot even produce the $x$ in the numerator.

connection

Partial fractions sits at a crossroads of algebra and calculus. It depends directly on Polynomial Factorization and the Fundamental Theorem of Algebra to guarantee that the denominator splits into linear and quadratic pieces. It is the gateway to Integration of Rational Functions, turning each piece into a logarithm or arctangent. In analysis it powers the method of Telescoping Series, where decomposing a term reveals massive cancellation. In engineering it is the inverse step of the Laplace Transform, converting transfer functions back into time-domain signals. It also relates to Residues in Complex Analysis, where the cover-up coefficients are precisely residues at poles.

🔬 Where this appears

⏳ History

- 1650s–1670s: Early integration of simple rational forms appears in the work of Mercator and others studying the area under hyperbolas (yielding logarithms).

- 1691–1700s: Leibniz and Johann Bernoulli develop the systematic decomposition of rational functions to integrate them, recognizing logs and arctangents as the resulting forms.

- 1702: Leibniz publishes work noting that integration of rational functions reduces to the cases of linear and quadratic denominators.

- 1740s: Euler refines and generalizes the technique, handling repeated factors and complex roots, in his foundational analysis texts.

- 1748: Euler's Introductio in analysin infinitorum presents partial fractions as a standard tool, including expansions of trigonometric and other functions.

- 1820s: Cauchy connects the coefficients to residues via complex analysis, giving the cover-up method a rigorous foundation.

- 1890s–1900s: Oliver Heaviside popularizes the operational/cover-up shortcut in electrical engineering for solving circuit equations.

- 1930s: The Laplace transform formalism standardizes partial fractions as the inversion technique in engineering mathematics.

open-problem

While partial fractions over the real numbers is fully solved, subtleties remain in practice and in theory. Factoring the denominator is the true bottleneck: by the Abel–Ruffini theorem, polynomials of degree five and higher have no general radical formula for their roots, so exact decomposition may be impossible in closed form and one must resort to numerical roots. Repeated and complex roots introduce numerical instability — coefficients computed by the cover-up method become ill-conditioned when roots cluster together. Over different fields (rationals vs. reals vs. complex numbers) the irreducible factors differ, so the "simplest" decomposition is field-dependent. Generalizing to functions of several variables (rational functions of $x$ and $y$ ) leads to deep questions in algebraic geometry where no simple universal decomposition exists.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also