# Partial Fraction Calculator

Decompose rational expressions into partial fractions. Handles linear numerators over distinct linear factors. Shows coefficients and verification.

## What this calculates

Break a rational expression into simpler partial fractions. Enter the numerator coefficients and the two roots of the denominator, and the calculator finds A and B so that (px + q) / ((x - r1)(x - r2)) = A/(x - r1) + B/(x - r2). For example, (5x + 3) / ((x - 1)(x + 2)) decomposes to 8/3/(x - 1) + 7/3/(x + 2).

## Inputs

- **Numerator: coefficient of x** — The coefficient of x in the numerator (for px + q).
- **Numerator: constant term** — The constant term in the numerator.
- **First root (r₁)** — First root of denominator, i.e., (x - r₁) is a factor.
- **Second root (r₂)** — Second root of denominator, i.e., (x - r₂) is a factor.

## Outputs

- **Original Expression** — formatted as text — The rational expression before decomposition.
- **A (first partial fraction)** — Coefficient A in A/(x - r₁).
- **B (second partial fraction)** — Coefficient B in B/(x - r₂).
- **Partial Fraction Decomposition** — formatted as text — The decomposed form: A/(x - r₁) + B/(x - r₂).
- **Verification** — formatted as text — Expanding back confirms the result matches the original.

## Details

**What Are Partial Fractions?**

Partial fraction decomposition rewrites a complex fraction as a sum of simpler fractions. It is one of the core techniques in calculus, especially for integration.

**The Method (Distinct Linear Factors)**

Given (px + q) / ((x - r1)(x - r2)), we want to find constants A and B such that:

(px + q) / ((x - r1)(x - r2)) = A/(x - r1) + B/(x - r2)

**How to Find A and B**

The cover-up method is the fastest approach:
- To find A, plug x = r1 into the numerator and divide by the remaining factor: A = (p*r1 + q) / (r1 - r2)
- To find B, plug x = r2: B = (p*r2 + q) / (r2 - r1)

**Worked Example**

Decompose (3x + 5) / ((x - 2)(x + 1)):
1. A = (3(2) + 5) / (2 - (-1)) = 11/3
2. B = (3(-1) + 5) / ((-1) - 2) = 2/(-3) = -2/3
3. Result: (11/3)/(x - 2) + (-2/3)/(x + 1)

**Why This Matters**

Partial fractions make otherwise difficult integrals easy to evaluate. For instance, integrating 1/((x-1)(x+1)) directly is hard, but integrating (1/2)/(x-1) - (1/2)/(x+1) is easy since each term gives a natural logarithm.

## Frequently Asked Questions

**Q: What is partial fraction decomposition?**

A: It is a technique that breaks a complex rational expression (a fraction where both numerator and denominator are polynomials) into a sum of simpler fractions. Each simpler fraction has a denominator that is a factor of the original denominator.

**Q: When do you use partial fractions?**

A: The most common use is in integral calculus, where decomposing a fraction makes it possible to integrate term by term. It also appears in solving differential equations, inverse Laplace transforms, and simplifying transfer functions in engineering.

**Q: What if the denominator has repeated roots?**

A: If a root repeats (e.g., (x - 2)^2), the decomposition includes terms with increasing powers: A/(x - 2) + B/(x - 2)^2. This calculator handles distinct linear factors only. For repeated or irreducible quadratic factors, additional terms are needed.

**Q: Does the numerator need to be lower degree than the denominator?**

A: Yes. Partial fraction decomposition requires the numerator degree to be strictly less than the denominator degree (a proper fraction). If it is not, you must first perform polynomial long division to extract the polynomial part.

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Source: https://vastcalc.com/calculators/math/partial-fraction
Category: Math
Last updated: 2026-04-08