Partial Fraction Decomposition Calculator

What Is Partial Fraction Decomposition?

Partial fraction decomposition (PFD) is an algebraic technique used to break down a complex rational expression $\frac{P(x)}{Q(x)}$ into a sum of simpler fractions with lower-degree denominators. It is the reverse operation of combining fractions with a common denominator.

This technique is indispensable in calculus for evaluating antiderivatives and indefinite integrals, in differential equations for computing inverse Laplace transforms, and in electrical engineering for transfer function analysis. For arithmetic fraction tools and polynomial operations, visit our Fractions Calculator, Fraction Simplifier, and Quadratic Calculator.

Conditions for Partial Fraction Decomposition

Before performing decomposition, the rational expression $\frac{P(x)}{Q(x)}$ must be a proper rational function, meaning the degree of the numerator $P(x)$ must be strictly less than the degree of the denominator $Q(x)$:

$$\deg(P) < \deg(Q)$$

If $\deg(P) \ge \deg(Q)$ (an improper fraction), you must first perform polynomial long division to express the fraction as a quotient polynomial plus a proper remainder fraction:

$$\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)}$$

The Four Types of Denominator Factors

The algebraic structure of the decomposition depends entirely on how the denominator $Q(x)$ factors over real numbers:

1. Distinct Linear Factors $(x - r_1)(x - r_2)\dots$

For each distinct non-repeated linear factor $(x - r)$, include a single term with a constant numerator:

$$\frac{P(x)}{(x - r_1)(x - r_2)} = \frac{A}{x - r_1} + \frac{B}{x - r_2}$$

2. Repeated Linear Factors $(x - r)^k$

For a linear factor $(x - r)$ raised to the power $k$, include $k$ successive terms with ascending powers:

$$\frac{P(x)}{(x - r)^k} = \frac{A_1}{x - r} + \frac{A_2}{(x - r)^2} + \dots + \frac{A_k}{(x - r)^k}$$

3. Irreducible Quadratic Factors $(ax^2 + bx + c)$

When a quadratic factor cannot be factored into real linear terms ($b^2 - 4ac < 0$), its numerator must be a linear expression $(Ax + B)$:

$$\frac{P(x)}{(x - r)(ax^2 + bx + c)} = \frac{A}{x - r} + \frac{Bx + C}{ax^2 + bx + c}$$

4. Repeated Irreducible Quadratic Factors $(ax^2 + bx + c)^k$

Similar to repeated linear factors, repeated quadratic factors generate ascending powers with linear numerators:

$$\frac{A_1 x + B_1}{ax^2 + bx + c} + \frac{A_2 x + B_2}{(ax^2 + bx + c)^2} + \dots + \frac{A_k x + B_k}{(ax^2 + bx + c)^k}$$

Methods for Solving Unknown Coefficients

Heaviside Cover-Up Method

For distinct linear factors, the Heaviside cover-up method provides an instant shortcut:

$$A = \lim_{x \to r_1} (x - r_1) \frac{P(x)}{Q(x)}$$

Simply cover the $(x - r_1)$ factor in the denominator and evaluate the remaining expression at $x = r_1$.

System of Linear Equations (Equating Coefficients)

Multiply both sides of the template by the common denominator, expand the polynomial on the right side, and equate corresponding powers of $x$ ($x^3, x^2, x^1, x^0$) to obtain a solvable system of linear equations.

Step-by-Step Worked Example

Decompose $\frac{3x + 5}{(x + 1)(x - 2)}$:

1. Set up partial fractions: $$\frac{3x + 5}{(x + 1)(x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2}$$
2. Clear denominators: $$3x + 5 = A(x - 2) + B(x + 1)$$
3. Substitute $x = -1$: $$3(-1) + 5 = A(-1 - 2) + 0 \implies 2 = -3A \implies A = -\frac{2}{3}$$
4. Substitute $x = 2$: $$3(2) + 5 = 0 + B(2 + 1) \implies 11 = 3B \implies B = \frac{11}{3}$$
5. Final Result: $$\frac{3x + 5}{(x + 1)(x - 2)} = -\frac{2/3}{x + 1} + \frac{11/3}{x - 2}$$