Partial Fraction Decomposition Calculator
Decompose rational expressions P(x)/Q(x) into partial fractions. Handles linear, repeated linear, and irreducible quadratic denominator factors with step-by-step coefficient solving.
Supports: 2x^2 + 3x - 1, 5x + 2, 7, etc.
What Is the Partial Fraction Decomposition Calculator?
Partial fraction decomposition rewrites a rational function P(x)/Q(x) as a sum of simpler fractions whose denominators are factors of Q(x). The result is much easier to integrate, differentiate, or use in inverse Laplace transforms. The technique is essential in calculus, differential equations, and control theory.
The Heaviside cover-up method finds coefficients for simple linear factors in one step: to find A for the term A/(x+a), multiply both sides by (x+a) and substitute x = −a, which “covers up” that denominator factor. For repeated factors and irreducible quadratics, coefficient matching solves a small linear system for the remaining unknowns.
Formula
Requirement: deg(P) < deg(Q) (proper fraction)
Case 1 — Distinct linear factors:
P(x)/[(x+a)(x+b)] = A/(x+a) + B/(x+b)
A = P(−a)/(b−a), B = P(−b)/(a−b) [Heaviside cover-up]
Case 2 — Repeated linear factor:
P(x)/(x+a)² = A/(x+a) + B/(x+a)²
B = P(−a), A = P'(−a)
Case 3 — Irreducible quadratic:
P(x)/[(x+a)(x²+bx+c)] = A/(x+a) + (Bx+C)/(x²+bx+c)
A = cover-up; B, C from coefficient matching
How to Use
- 1Select template: Choose the denominator form that matches your rational expression (e.g., (x+a)(x+b) for two distinct linear factors).
- 2Enter numerator: Type the numerator polynomial using x^2, x, and constants. Degree must be less than the denominator degree.
- 3Enter factor values: Supply the numeric values for a, b, c in the denominator factors.
- 4Decompose: Click the button. Results show each partial fraction coefficient and the full step-by-step working.
- 5Improper fractions: If the numerator degree ≥ denominator degree, first perform polynomial long division and decompose the remainder.
Example Calculation
Example: (3x + 5) / [(x+1)(x+2)]
- ›Template: (x+a)(x+b) with a=1, b=2; Numerator: 3x + 5
- ›A = P(−1)/(2−1) = (−3+5)/1 = 2/1 = 2
- ›B = P(−2)/(1−2) = (−6+5)/(−1) = −1/(−1) = 1
- ›Result: (3x+5)/[(x+1)(x+2)] = 2/(x+1) + 1/(x+2)
- ›Verify: 2(x+2) + 1(x+1) = 2x+4+x+1 = 3x+5 ✓
Calculus application: ∫(3x+5)/[(x+1)(x+2)]dx
- ›After decomposition: ∫[2/(x+1) + 1/(x+2)]dx
- ›= 2·ln|x+1| + ln|x+2| + C
- ›Much simpler than integrating the original rational function directly.
Understanding Partial Fraction Decomposition
Common Denominator Templates
| Denominator | Partial fraction form | Method |
|---|---|---|
| (x+a)(x+b) | A/(x+a) + B/(x+b) | Cover-up at x=−a and x=−b |
| (x+a)² | A/(x+a) + B/(x+a)² | Cover-up + differentiation |
| (x+a)(x+b)(x+c) | A/(x+a)+B/(x+b)+C/(x+c) | Three cover-ups |
| (x+a)(x²+bx+c) | A/(x+a) + (Bx+C)/(x²+bx+c) | Cover-up + coefficient match |
| (x+a)²(x+b) | A/(x+a)+B/(x+a)²+C/(x+b) | Cover-up + coefficient match |
Integration After Decomposition
Each partial fraction integrates to a known form: ∫A/(x+a)dx = A·ln|x+a| + C for linear factors, ∫B/(x+a)²dx = −B/(x+a) + C for repeated factors, and the quadratic form integrates to arctan plus a logarithm. Decomposition converts one difficult integral into several trivial ones.
Frequently Asked Questions
What does "proper fraction" mean and why is it required?
Proper means deg(P) < deg(Q). Partial fractions only apply to proper rational functions. For improper fractions (deg(P) ≥ deg(Q)), use polynomial long division first to get a polynomial plus a proper remainder, then decompose the remainder.
What is the Heaviside cover-up method?
For A/(x+a) with a non-repeated linear factor, multiply the whole identity by (x+a) and set x = −a. All terms except A vanish, giving A = P(−a) divided by the product of remaining denominator factors evaluated at x = −a. Fast, one-step method for simple poles.
Why does a repeated factor (x+a)² require two terms?
A double pole requires two terms to be algebraically complete: A/(x+a) captures the simple-pole behavior and B/(x+a)² captures the double-pole behavior. Together they can reproduce any numerator of degree < 2 over that denominator.
When do I use the (Bx+C)/(x²+bx+c) form?
Use (Bx+C)/(x²+bx+c) when the quadratic has no real roots (discriminant b²−4c < 0). It cannot be split into linear factors over ℝ, so the partial fraction contribution must have a linear numerator. This typically leads to arctangent terms when integrated.
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