Partial Fraction Decomposition Calculator
Decompose rational expressions into partial fractions for integration and analysis
Enter polynomial expression
Enter in factored form
Understanding Partial Fraction Decomposition
What is Partial Fraction Decomposition?
Partial fraction decomposition is the reverse of adding fractions. It breaks a complex rational expression into a sum of simpler fractions, which is especially useful for integration in calculus.
Three Main Cases
Case 1: Linear Distinct Factors
For each factor (ax + b), use A/(ax + b)
Example: 1/[(x-1)(x+2)] = A/(x-1) + B/(x+2)
Case 2: Repeated Linear Factors
For (ax + b)ⁿ, use A/(ax+b) + B/(ax+b)² + ... + N/(ax+b)ⁿ
Example: 1/(x-1)³ = A/(x-1) + B/(x-1)² + C/(x-1)³
Case 3: Irreducible Quadratic Factors
For (ax² + bx + c), use (Ax + B)/(ax² + bx + c)
Example: 1/[(x-1)(x²+1)] = A/(x-1) + (Bx+C)/(x²+1)
Solution Methods
- Substitution Method: Substitute convenient values of x to solve for constants
- Equating Coefficients: Expand and match coefficients of like powers
- Cover-Up Method: Quick technique for linear distinct factors
Complete Example
Decompose: (2x + 1)/[(x - 1)(x + 2)]
- Set up: (2x + 1)/[(x-1)(x+2)] = A/(x-1) + B/(x+2)
- Multiply by LCD: 2x + 1 = A(x+2) + B(x-1)
- Let x = 1: 3 = 3A → A = 1
- Let x = -2: -3 = -3B → B = 1
- Result: 1/(x-1) + 1/(x+2)
Frequently Asked Questions
When do I need partial fraction decomposition?
Primarily in calculus for integrating rational functions. It's also used in differential equations, Laplace transforms, and signal processing.
What if the numerator degree ≥ denominator degree?
First perform polynomial long division to get a polynomial plus a proper fraction. Then decompose the proper fraction part.
How do I know if a quadratic is irreducible?
Check the discriminant (b² - 4ac). If it's negative, the quadratic doesn't factor over real numbers and is irreducible.
Can I use partial fractions for all rational expressions?
Yes, as long as the denominator can be factored (possibly with complex numbers). The denominator must be non-zero.
Why does the quadratic case need (Ax + B) not just A?
A quadratic denominator has degree 2, so the numerator needs degree 1 (thus two constants A and B) to handle all possible cases.
How many constants will I have?
The number of unknown constants equals the degree of the denominator. Degree 2 denominator = 2 constants, etc.
Can partial fractions be added back to get the original?
Yes! That's a great way to check your work. Add your partial fractions and simplify - you should get the original expression.
What's the cover-up method?
A shortcut for linear distinct factors: 'cover up' a factor in the denominator, substitute its zero into what remains. This gives you that constant directly.