The Ultimate Guide to Partial Fraction Decomposition
Everything you need to know about decomposing rational functions for Calculus and Algebra.
Partial Fraction Decomposition (PFD) is one of the most powerful algebraic techniques available to mathematics students. While it is often introduced in Pre-Calculus or Algebra II, its true utility shines in Calculus (for integration) and Differential Equations (for Laplace Transforms). This guide will walk you through the theory, the methods, and the common pitfalls.
1. What is Partial Fraction Decomposition?
At its core, PFD is the reverse process of adding fractions. When you learn algebra, you are often asked to combine two fractions into one by finding a common denominator:
$$ \frac{2}{x-1} + \frac{3}{x+2} = \frac{2(x+2) + 3(x-1)}{(x-1)(x+2)} = \frac{5x+1}{x^2+x-2} $$
Partial Fraction Decomposition asks the opposite question: If I start with $\frac{5x+1}{x^2+x-2}$, how do I break it apart into simpler fractions?
We do this because the "simpler fractions" are much easier to work with. For example, integrating $\frac{5x+1}{x^2+x-2}$ is difficult directly, but integrating $\frac{2}{x-1} + \frac{3}{x+2}$ is a trivial matter of natural logarithms.
2. Prerequisites: When can you use it?
Before applying PFD, you must ensure the rational function is proper. A rational function $\frac{P(x)}{Q(x)}$ is proper if the degree of the numerator $P(x)$ is strictly less than the degree of the denominator $Q(x)$.
- Proper: $\frac{x}{x^2+1}$ (Degree 1 < Degree 2)
- Improper: $\frac{x^3}{x^2+1}$ (Degree 3 > Degree 2)
If you encounter an improper fraction, you must first perform Polynomial Long Division. The PFD technique is then applied only to the remainder term.
3. The Four Cases of Decomposition
The structure of your decomposition depends entirely on how the denominator factors. We generally categorize these into four distinct cases.
Case 1: Distinct Linear Factors
If the denominator factors into unique linear terms like $(x-a)(x-b)(x-c)$, the decomposition takes the form:
$$ \frac{P(x)}{Q(x)} = \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c} $$
This is the most common scenario found in introductory calculus problems.
Case 2: Repeated Linear Factors
If a linear factor is repeated, such as $(x-a)^3$, you must include a term for every power from 1 up to the multiplicity. It is a common mistake to only include the highest power.
$$ \frac{P(x)}{(x-a)^3} = \frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{(x-a)^3} $$
Case 3: Distinct Irreducible Quadratic Factors
Sometimes a quadratic term like $(x^2+1)$ or $(x^2+4)$ cannot be factored further using real numbers. In this case, the numerator must be a linear term ($Ax+B$) rather than a constant.
$$ \frac{P(x)}{(x-a)(x^2+1)} = \frac{A}{x-a} + \frac{Bx+C}{x^2+1} $$
Case 4: Repeated Quadratic Factors
Similar to repeated linear factors, if a quadratic is repeated, you build up the powers, keeping the linear numerator ($Ax+B$) for each.
$$ \frac{P(x)}{(x^2+1)^2} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2} $$
4. Solving Methods: How to find A, B, and C
Once you have set up the general form, you need to solve for the unknown constants. There are two primary methods for this.
Method A: Equating Coefficients
This is the systematic "brute force" method.
1. Multiply the entire equation by the common denominator to clear the fractions.
2. Expand all polynomials on the right side.
3. Group terms by powers of $x$ (e.g., all $x^2$ terms, all $x$ terms).
4. Set up a system of linear equations by matching coefficients from the left side to the right side.
Method B: The Heaviside "Cover-Up" Method
This is a faster trick, primarily useful for distinct linear factors.
1. Multiply by the denominator to clear fractions.
2. Instead of expanding, substitute convenient values for $x$ that make terms disappear.
Example: If you have $A(x+2) + B(x-1)$, setting $x=1$ eliminates $A$, allowing you to solve for $B$ instantly.
5. Worked Example
Let's decompose $\frac{7x+13}{x^2+2x-3}$.
Step 1: Factor the denominator.
$x^2+2x-3 = (x-1)(x+3)$.
Step 2: Set up the form.
$$ \frac{7x+13}{(x-1)(x+3)} = \frac{A}{x-1} + \frac{B}{x+3} $$
Step 3: Clear the fractions.
Multiply by $(x-1)(x+3)$:
$$ 7x+13 = A(x+3) + B(x-1) $$
Step 4: Solve for constants (Substitution Method).
Let $x=1$:
$7(1)+13 = A(1+3) + B(0)$
$20 = 4A \implies \mathbf{A=5}$
Let $x=-3$:
$7(-3)+13 = A(0) + B(-3-1)$
$-8 = -4B \implies \mathbf{B=2}$
Final Answer:
$$ \frac{5}{x-1} + \frac{2}{x+3} $$
6. Applications in Calculus and Engineering
Why do we do this? In Calculus II, integrating $\int \frac{1}{1-x^2}dx$ is standard. By decomposing it into $\frac{1}{2}(\frac{1}{1-x} + \frac{1}{1+x})$, the solution becomes logarithmic.
In Control Systems Engineering, the Inverse Laplace Transform is essential for analyzing system stability. The Laplace domain functions are almost always rational expressions that require partial fraction decomposition to return to the time domain.
Conclusion
Mastering Partial Fraction Decomposition is a rite of passage for STEM students. Whether you use the Heaviside cover-up method for speed or equate coefficients for complex quadratic factors, the goal remains the same: simplify to conquer. Use the calculator at the top of this page to check your algebra, but always practice the manual steps to ensure deep understanding.