Partial Fraction Decomposition
Decompose (ax+b)/((x-r₁)(x-r₂)) into A/(x-r₁) + B/(x-r₂). For repeated roots: A/(x-r) + B/(x-r)². For irreducible quadratics: (Ax+B)/(x²+c). Essential for integration and Laplace transforms.
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Cover-up method: multiply by (x-r₁), plug x=r₁ to get A directly — no system needed. Repeated roots require higher powers: (x-r)² gets A/(x-r) + B/(x-r)². Irreducible quadratics (x²+c) need numerator Ax+B, not just a constant.
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Why: Partial fraction decomposition is essential for integrating rational functions and solving differential equations via Laplace transforms. Control theory and signal processing rely on it.
How: Distinct linear: multiply by denominator, plug roots to get A, B. Repeated: form A/(x-r)+B/(x-r)², solve system. Quadratic: (Ax+B)/(x²+c). Cover-up method: plug each root to isolate one constant.
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📐 Examples — Click to Load
Decomposition Type
Partial Fraction Constants
Contribution of Each Fraction
📐 Calculation Steps
For educational and informational purposes only. Verify with a qualified professional.
🧮 Fascinating Math Facts
Partial fractions turn integrals of rational functions into sums of log and arctan terms.
— Calculus
Cover-up: for A/(x-1)+B/(x+2), plug x=1 to get A, x=-2 to get B.
— Method
📋 Key Takeaways
- • Partial fraction decomposition breaks a rational function into simpler fractions
- • Distinct linear factors: (ax+b)/((x-r₁)(x-r₂)) = A/(x-r₁) + B/(x-r₂)
- • Repeated roots: (ax+b)/(x-r)² = A/(x-r) + B/(x-r)²
- • Irreducible quadratics: Use form (Ax+B)/(ax²+bx+c)
- • Cover-up method: Substitute roots to find constants quickly
💡 Why Partial Fractions Matter
📖 Method of Undetermined Coefficients
For (3x+5)/((x-1)(x+2)) = A/(x-1) + B/(x+2): Multiply both sides by (x-1)(x+2) to get 3x+5 = A(x+2) + B(x-1). Substitute x=1: 8=3A ⇒ A=8/3. Substitute x=-2: -1=-3B ⇒ B=1/3.
Cover-Up Method
To find A for A/(x-1): "Cover up" (x-1) in the original, substitute x=1 into what remains: A = (3(1)+5)/(1+2) = 8/3.
📊 Decomposition Forms
| Denominator | Partial Fraction Form |
|---|---|
| (x-r₁)(x-r₂) | A/(x-r₁) + B/(x-r₂) |
| (x-r)² | A/(x-r) + B/(x-r)² |
| x²+1 (irreducible) | (Ax+B)/(x²+1) |
| (x-r)(x²+1) | A/(x-r) + (Bx+C)/(x²+1) |
🎯 Expert Tips
💡 Check Proper First
If deg(P) ≥ deg(Q), do polynomial long division first. Decompose only the remainder over Q.
💡 Verify Your Answer
Add the partial fractions back together. You should get the original rational expression.
💡 Laplace Applications
For L⁻¹Y(s), decompose Y(s) into terms like 1/(s-a) whose inverse is e^(at).
💡 Integration Result
∫A/(x-r)dx = A·ln|x-r|. ∫(Ax+B)/(x²+1)dx uses arctan. Each term has a standard antiderivative.
❓ FAQ
What is partial fraction decomposition?
Breaking a rational function P(x)/Q(x) into a sum of simpler fractions whose denominators are factors of Q(x). Essential for integration and Laplace transforms.
When do I need polynomial division first?
When the degree of the numerator is greater than or equal to the degree of the denominator. The result is polynomial + proper fraction.
What is the cover-up method?
For distinct linear factors: to find A in A/(x-r), substitute x=r into the original after "covering" (x-r). A = P(r)/Q(r) where Q is the product of other factors evaluated at r.
How do I handle repeated roots?
For (x-r)ⁿ, include terms A₁/(x-r) + A₂/(x-r)² + ... + Aₙ/(x-r)ⁿ. Substitute x=r for the highest power term, then equate coefficients for others.
What about irreducible quadratics?
Use (Ax+B)/(ax²+bx+c). Multiply through, expand, and equate coefficients of 1, x, x² to get a system of equations.
Why is this useful in calculus?
∫1/(x-a)dx = ln|x-a|. ∫1/(x²+a²)dx = (1/a)arctan(x/a). Decomposition turns hard integrals into sums of easy ones.
⚠️ Disclaimer: This calculator handles common decomposition types. For complex denominators or improper fractions, manual polynomial division may be required first. Educational use only.
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