Partial fraction decomposition is a technique usedto break down a rational function into simpler fractions that are easier to integrate or differentiate, and understanding how to find the partial fraction decomposition of the rational function is essential for solving many calculus problems.
Introduction
A rational function is any expression that can be written as the quotient of two polynomials, ( \frac{P(x)}{Q(x)} ), where (P(x)) and (Q(x)) have no common factors other than constants. Still, when the denominator (Q(x)) can be factored into linear or irreducible quadratic terms, the fraction can often be expressed as a sum of simpler fractions known as partial fractions. This transformation simplifies tasks such as integration, differentiation, and solving differential equations.
What is a Rational Function?
Definition and General Form A rational function is defined as
[R(x)=\frac{P(x)}{Q(x)}, ]
where (P(x)) and (Q(x)) are polynomials and (Q(x)\neq 0). Here's the thing — ### Proper vs. Now, if the degree of (P(x)) is greater than or equal to the degree of (Q(x)), polynomial long division is performed first to rewrite the function as a polynomial plus a proper rational function (one where the numerator’s degree is lower than the denominator’s). Worth adding: improper Fractions - Proper rational function: (\deg(P) < \deg(Q)). So the degree of the numerator may be less than, equal to, or greater than the degree of the denominator. - Improper rational function: (\deg(P) \ge \deg(Q)) Worth keeping that in mind..
Only proper fractions can be directly decomposed into partial fractions without prior division.
Basic Principles of Partial Fraction Decomposition
Linear Factors
If (Q(x)) factors completely into distinct linear terms,
[ Q(x) = (x-a_1)(x-a_2)\dots (x-a_n), ]
the decomposition takes the form
[ \frac{P(x)}{Q(x)} = \frac{A_1}{x-a_1} + \frac{A_2}{x-a_2} + \dots + \frac{A_n}{x-a_n}, ]
where (A_1, A_2, \dots, A_n) are constants to be determined.
Repeated Linear Factors
When a linear factor repeats, say ((x-a)^k), the corresponding terms are
[ \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \dots + \frac{A_k}{(x-a)^k}. ]
Irreducible Quadratic Factors
If (Q(x)) contains an irreducible quadratic factor ( (x^2+bx+c) ), the associated term is
[ \frac{Bx+C}{x^2+bx+c}, ]
with (B) and (C) constants. For a repeated quadratic factor, additional powers are added similarly Not complicated — just consistent..
Step‑by‑Step Procedure
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Factor the Denominator
- Write (Q(x)) as a product of its irreducible linear and quadratic factors.
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Set Up the Partial Fraction Form
- Assign an unknown constant (or linear polynomial for quadratics) for each factor according to the rules above.
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Clear the Denominator
- Multiply both sides of the equation by the full denominator (Q(x)) to eliminate fractions.
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Expand and Collect Like Terms
- Distribute the unknown constants and combine like powers of (x).
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Equate Coefficients
- Match the coefficients of corresponding powers of (x) on both sides of the equation to create a system of linear equations.
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Solve the System
- Use substitution, elimination, or matrix methods to find the values of the unknown constants.
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Write the Final Decomposition
- Substitute the solved constants back into the partial fraction form. ### Example
Find the partial fraction decomposition of
[ \frac{2x^2+3x+1}{(x-1)(x+2)^2}. ]
Step 1: The denominator is already factored: ((x-1)(x+2)^2) The details matter here..
Step 2: Set up the form
[ \frac{2x^2+3x+1}{(x-1)(x+2)^2}= \frac{A}{x-1}+ \frac{B}{x+2}+ \frac{C}{(x+2)^2}. ]
Step 3: Multiply by ((x-1)(x+2)^2)
[ 2x^2+3x+1 = A(x+2)^2 + B(x-1)(x+2) + C(x-1). ]
Step 4: Expand
[ \begin{aligned} A(x^2+4x+4) &+ B(x^2 + x -2) + C(x-1) \ &= (A+B)x^2 + (4A+B+C)x + (4A-2B-C). \end{aligned} ]
Step 5: Equate coefficients with (2x^2+3x+1):
[ \begin{cases} A + B = 2,\ 4A + B + C = 3,\ 4A - 2B - C = 1. \end{cases} ]
Step 6: Solve
From the first equation, (B = 2 - A). Substitute into the other two:
[ \begin{aligned} 4A + (2-A) + C &= 3 ;\Rightarrow; 3A + C = 1,\ 4A - 2(2-A) - C &= 1 ;\Rightarrow; 6A - 4 - C = 1 ;\Rightarrow; 6A - C = 5. \end{aligned} ]
Add the two equations: (9A = 6 \Rightarrow A = \frac{2}{3}).
Still, then (B = 2 - \frac{2}{3} = \frac{4}{3}). Finally, (3A + C = 1 \Rightarrow 2 + C = 1 \Rightarrow C = -1).
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