Partial Fraction Decomposition Calculator

Decompose a Rational Function into Partial Fractions

Enter a rational function P(x)/Q(x) (numerator degree < denominator degree) and click Decompose. The calculator provides step-by-step partial fraction decomposition.

Decomposition result will appear here.

Note: This calculator handles simple linear factors. For repeated or quadratic factors, follow the examples below.

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Partial Fraction Decomposition – Complete Explanation

Partial fraction decomposition expresses a rational function as a sum of simpler fractions, making integration and algebraic manipulation easier. This method is essential in calculus, differential equations, and engineering.

Partial Fraction Principle

For a proper rational function P(x)/Q(x), if Q(x) factors into linear factors, we can write:

P(x)/Q(x) = A/(factor1) + B/(factor2) + ...

Constants A, B, etc., are determined by solving equations.

1. Steps for Decomposition

1. Check degree: Ensure numerator degree < denominator degree. If not, perform polynomial division first.
2. Factor the denominator Q(x) completely into linear and/or irreducible quadratic factors.
3. Set up the form: Write a sum of fractions with unknown constants (A, B, C, ...) based on factor types.
4. Clear denominators: Multiply both sides by Q(x) to eliminate fractions.
5. Solve for constants using substitution (plug in roots) or equating coefficients.
6. Write the final decomposition.

2. Step-by-Step Example

Example 1: Distinct Linear Factors

Decompose: (2x+3)/(x^2 - x - 2)

Step 1: Factor denominator: x^2 - x - 2 = (x-2)(x+1)

Step 2: Set up: (2x+3)/((x-2)(x+1)) = A/(x-2) + B/(x+1)

Step 3: Multiply both sides by (x-2)(x+1):
2x + 3 = A(x+1) + B(x-2)

Step 4: Solve for constants:
        x = 2 → 2*2 + 3 = 7 = A(3) + B(0) → A = 7/3
        x = -1 → 2*(-1)+3 = 1 = A(0) + B(-3) → B = -1/3

Step 5: Final: (2x+3)/(x^2-x-2) = 7/3 * 1/(x-2) - 1/3 * 1/(x+1)

3. More Examples

Example 1: (x+5)/(x²+3x+2)

Factor: (x+1)(x+2) → Setup: A/(x+1) + B/(x+2)

Multiply: x+5 = A(x+2) + B(x+1)

Solve: x=-1 → 4 = A(1) → A=4; x=-2 → 3 = B(-1) → B=-3

Final: (x+5)/(x²+3x+2) = 4/(x+1) - 3/(x+2)

Example 2: (3x+7)/(x²-4)

Factor: (x-2)(x+2) → Setup: A/(x-2) + B/(x+2)

Multiply: 3x+7 = A(x+2) + B(x-2)

Solve: x=2 → 13 = 4A → A=13/4; x=-2 → 1 = -4B → B=-1/4

Final: (3x+7)/(x²-4) = 13/4·1/(x-2) - 1/4·1/(x+2)

Example 3: (x²+3x+2)/(x³-x)

Factor denominator: x(x-1)(x+1)

Setup: A/x + B/(x-1) + C/(x+1)

Multiply: x²+3x+2 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1)

Solve: x=0 → 2 = A(-1)(1) → A=-2; x=1 → 6 = B(1)(2) → B=3; x=-1 → 0 = C(-1)(-2) → C=0

Final: -2/x + 3/(x-1)

4. Case 2: Repeated Linear Factors

If a factor (x-a) appears k times, include terms: A₁/(x-a) + A₂/(x-a)² + ... + Aₖ/(x-a)ᵏ

Example: Repeated Linear Factor

Decompose: (x+1)/(x²(x-2))

Setup: A/x + B/x² + C/(x-2)

Multiply: x+1 = A x(x-2) + B(x-2) + C x²

Solve: x=0 → 1 = -2B → B = -1/2; x=2 → 3 = 4C → C = 3/4; Compare x²: 0 = A + C → A = -3/4

Final: -3/(4x) - 1/(2x²) + 3/(4(x-2))

5. Case 3: Irreducible Quadratic Factors

For an irreducible quadratic factor ax²+bx+c, use term (Ax+B)/(ax²+bx+c).

Example: Quadratic Factor

Decompose: (2x²+3x+1)/((x²+1)(x-1))

Setup: (Ax+B)/(x²+1) + C/(x-1)

Multiply: 2x²+3x+1 = (Ax+B)(x-1) + C(x²+1)

Expand: = Ax² - Ax + Bx - B + Cx² + C = (A+C)x² + (-A+B)x + (-B+C)

Compare coefficients:
x²: A+C = 2
x¹: -A+B = 3
x⁰: -B+C = 1

Solve: From 1st: C=2-A; 3rd: -B+2-A=1 → -A-B=-1 → A+B=1
2nd: -A+B=3. Add to A+B=1 → 2B=4 → B=2, then A=-1, C=3

Final: (-x+2)/(x²+1) + 3/(x-1)

6. Case 4: Repeated Quadratic Factors

For (ax²+bx+c)ᵏ, include: (A₁x+B₁)/(ax²+bx+c) + (A₂x+B₂)/(ax²+bx+c)² + ...

7. Common Mistakes to Avoid

• Not factoring denominator completely (always factor over reals first)
• Forgetting proper setup for repeated or quadratic factors
• Arithmetic errors when solving for constants
• Not checking degree – improper fractions need long division first
• Sign errors in substitution method
• Missing terms when equating coefficients

8. Connection to the Remainder Theorem

The Remainder Theorem helps check factors quickly. If P(a)=0, then (x-a) is a factor of the denominator, which is essential for decomposition.

The Factor Theorem in Partial Fractions

Finding roots of Q(x) using P(a)=0 confirms linear factors and helps set up the decomposition form correctly.

9. Real-World Applications

• Calculus Integration: Integrating rational functions becomes straightforward after decomposition
• Differential Equations: Solving inverse Laplace transforms in engineering
• Signal Processing: Breaking transfer functions into simpler components
• Control Systems: Analyzing system response using partial fractions
• Physics: Solving differential equations in circuit analysis and mechanics
• Numerical Analysis: Rational function approximation

10. Practice Problems with Solutions

Problem 1: Decompose (7x-4)/(x²-x-6)

Solution: Factor: (x-3)(x+2). Setup: A/(x-3) + B/(x+2). Solve: A=17/5, B=18/5 → 17/5·1/(x-3) + 18/5·1/(x+2)

Problem 2: Decompose (x²+1)/(x(x-1)²)

Solution: Setup: A/x + B/(x-1) + C/(x-1)². Solve: A=1, B=0, C=2 → 1/x + 2/(x-1)²

Problem 3: Decompose (3x²+2x+1)/((x²+1)(x+2))

Solution: Setup: (Ax+B)/(x²+1) + C/(x+2). Solve: A=1, B=1, C=2 → (x+1)/(x²+1) + 2/(x+2)

Problem 4: Decompose (5x+3)/(x³-2x²+x-2)

Solution: Factor by grouping: (x-2)(x²+1). Setup: A/(x-2) + (Bx+C)/(x²+1). Solve: A=13/5, B=-13/5, C=1/5

11. Tips for Mastering Partial Fractions

• Always factor denominator completely before setting up
• Practice each case separately (linear, repeated, quadratic)
• Use both substitution and coefficient comparison to verify answers
• Check your work by combining the partial fractions back
• Remember: improper fractions require long division first
• When using substitution, choose values that make terms zero

12. Connection to Calculus: Integration

Once decomposed, integration becomes simple:

∫ A/(x-a) dx = A·ln|x-a| + C
∫ (Ax+B)/(x²+px+q) dx = use substitution or arctan forms
∫ A/(x-a)ⁿ dx = A·(x-a)⁻ⁿ⁺¹/(-n+1) + C

Calculus Example: ∫ (2x+3)/(x²-x-2) dx

After decomposition: ∫ [7/3·1/(x-2) - 1/3·1/(x+1)] dx

= 7/3·ln|x-2| - 1/3·ln|x+1| + C

13. Final Thoughts

Partial fraction decomposition is a fundamental algebraic technique that transforms complex rational functions into manageable pieces. It bridges algebra and calculus, enabling solutions that would otherwise be extremely difficult.

Use this calculator to verify your work, but practice manual decomposition to build confidence. With consistent practice, you'll recognize the correct form instantly and solve for constants efficiently.