What is partial fraction decomposition used for?

Partial fraction decomposition breaks a rational function into a sum of simpler fractions that are easier to integrate. It is primarily used in integral calculus but also appears in Laplace transforms and solving differential equations.

How do I set up partial fractions for distinct linear factors?

For each distinct linear factor (ax + b) in the denominator, write a term A/(ax + b). For repeated linear factors (ax + b)ⁿ, write A₁/(ax+b) + A₂/(ax+b)² + ... + Aₙ/(ax+b)ⁿ. Then solve for constants by matching coefficients.

What should I do before applying partial fraction decomposition?

First ensure the rational function is proper — the degree of the numerator must be less than the degree of the denominator. If the function is improper, perform polynomial long division first, then decompose the proper remainder part.

AI-Math - Partial Fractions Decomposition: The Complete Workflow

Partial fraction decomposition is the algebraic skill that lets you integrate any rational function on the planet. Instead of fighting one ugly fraction, you split it into pieces that are easy to integrate term by term. This guide walks through every case you'll meet.

The setup

A rational function is $\frac{P(x)}{Q(x)}$ where $P, Q$ are polynomials. Partial fractions only works when the degree of $P$ < degree of $Q$ . If not, do polynomial long division first to peel off the polynomial part.

After dividing, factor $Q(x)$ completely over the reals. Every factor falls into one of four categories.

The four cases

Case 1: distinct linear factors

If $Q(x) = (x - a)(x - b)$ , write:

$\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$

Example. Decompose $\frac{5x - 1}{(x - 1)(x + 2)}$ .

Multiply through: $5x - 1 = A(x + 2) + B(x - 1)$ .

Plug $x = 1$ : $4 = 3A \Rightarrow A = 4/3$ .
Plug $x = -2$ : $-11 = -3B \Rightarrow B = 11/3$ .

So $\frac{5x-1}{(x-1)(x+2)} = \frac{4/3}{x-1} + \frac{11/3}{x+2}$ .

Case 2: repeated linear factor

For $(x - a)^k$ , you need one term per power up to $k$ :

$\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \dots + \frac{A_k}{(x-a)^k}$

Case 3: irreducible quadratic factor

For each irreducible $x^2 + bx + c$ , use a numerator with two unknowns:

$\frac{Bx + C}{x^2 + bx + c}$

Case 4: repeated irreducible quadratic

Same idea as case 2, but each power gets a $Bx + C$ form.

Integration application

Once decomposed, integrate term by term:

$\int \frac{1}{x - a} dx = \ln|x - a| + C$
$\int \frac{1}{(x - a)^k} dx = \frac{-1}{(k-1)(x-a)^{k-1}} + C$ for $k > 1$
$\int \frac{Bx + C}{x^2 + bx + c} dx$ splits into a $\ln$ part and an $\arctan$ part.

Common mistakes

Forgetting to do long division first when degree of $P$ ≥ degree of $Q$ .
Skipping repeated terms — $(x - 1)^3$ requires three separate fractions.
Trying to factor irreducible quadratics — check the discriminant before forcing real roots.

Try with the AI Integral Solver

The Integral Solver automatically does partial fraction decomposition when needed and shows every step.

Related references:

Factor Calculator — for breaking $Q(x)$ apart
Polynomial Calculator — for the long-division setup
Limit Calculator — used in some PFD verification tricks

Partial Fractions Decomposition: The Complete Workflow

A no-fluff walkthrough of partial fractions — the four cases (distinct linear, repeated linear, irreducible quadratic, repeated quadratic) with worked examples and integration tips.