Partial fractions is a technique used in algebra, particularly in integral calculus, to break down a complex rational function into simpler fractions called partial fractions. This is useful for integrating more complicated functions and solving equations. The primary application is in the integration of rational functions, where one polynomial is divided by another.
General Concept
Given a rational function of the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials and the degree of \( P(x) \) is less than the degree of \( Q(x) \), the goal is to express \( \frac{P(x)}{Q(x)} \) as a sum of simpler fractions:
\[ \frac{P(x)}{Q(x)} = \sum \frac{A_i}{f_i(x)} \]
where \( f_i(x) \) are factors of the denominator \( Q(x) \).
Types of Partial Fractions
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Distinct Linear Factors: If the denominator \( Q(x) \) can be factored into distinct linear factors (like \( x - a \)), the form would be:
\[ \frac{P(x)}{Q(x)} = \frac{A_1}{x - a_1} + \frac{A_2}{x - a_2} + \ldots + \frac{A_n}{x - a_n} \]
Example: \[ \frac{2x + 3}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2} \]
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Repeated Linear Factors: If the denominator has repeated linear factors (like \( (x - a)^k \)), the form would be:
\[ \frac{P(x)}{Q(x)} = \frac{A_1}{x - a} + \frac{A_2}{(x - a)^2} + \ldots + \frac{A_k}{(x - a)^k} \]
Example: \[ \frac{3x + 5}{(x - 2)^2} = \frac{A}{x - 2} + \frac{B}{(x - 2)^2} \]
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Distinct Quadratic Factors: If the denominator contains distinct irreducible quadratic factors (like \( ax^2 + bx + c \)), the form would be:
\[ \frac{P(x)}{Q(x)} = \frac{Ax + B}{ax^2 + bx + c} \]
Example: \[ \frac{4x + 2}{(x^2 + 1)(x - 3)} = \frac{Ax + B}{x^2 + 1} + \frac{C}{x - 3} \]
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Repeated Quadratic Factors: If the denominator has repeated irreducible quadratic factors, the form would be:
\[ \frac{P(x)}{Q(x)} = \frac{Ax + B}{(ax^2 + bx + c)} + \frac{C}{(ax^2 + bx + c)^2} \]
Example: \[ \frac{3x + 4}{(x^2 + 1)^2} = \frac{Ax + B}{x^2 + 1} + \frac{C}{(x^2 + 1)^2} \]
Steps to Find Partial Fraction Decomposition
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Factor the Denominator: Factor \( Q(x) \) into linear and/or quadratic factors.
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Set Up the Decomposition: Write down the partial fraction decomposition form based on the types of factors—either distinct or repeated.
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Multiply Through by the Denominator: Eliminate the fractions by multiplying both sides by \( Q(x) \).
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Collect Like Terms: Expand the right-hand side and collect like terms.
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Equate Coefficients: Set the coefficients of corresponding powers of \( x \) equal to obtain a system of equations.
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Solve for Constants: Solve the system of equations to find the values of the constants in the partial fractions.
Example of Partial Fraction Decomposition
Let’s take a concrete example to see how this works. Consider:
\[ \frac{3x + 5}{(x - 2)(x + 1)} \]
Step 1: Set Up the Decomposition
This rational function can be expressed as:
\[ \frac{3x + 5}{(x - 2)(x + 1)} = \frac{A}{x - 2} + \frac{B}{x + 1} \]
Step 2: Multiply Through by the Denominator
Multiplying both sides by \( (x - 2)(x + 1) \) gives:
\[ 3x + 5 = A(x + 1) + B(x - 2) \]
Step 3: Expand and Collect Terms
Expanding the right side:
\[ 3x + 5 = Ax + A + Bx - 2B = (A + B)x + (A - 2B) \]
Step 4: Equate Coefficients
Setting coefficients equal gives us the system:
\[ A + B = 3 \quad (1) \] \[ A - 2B = 5 \quad (2) \]
Step 5: Solve for Constants
Now, we can solve this system:
From equation (1):
\[ A = 3 - B \]
Substituting \( A \) in equation (2):
\[ (3 - B) - 2B = 5 \implies 3 - 3B = 5 \implies -3B = 2 \implies B = -\frac{2}{3} \]
Substituting \( B \) back into equation (1):
\[ A - \frac{2}{3} = 3 \implies A = 3 + \frac{2}{3} = \frac{9}{3} + \frac{2}{3} = \frac{11}{3} \]
Thus, the partial fractions are:
\[ \frac{3x + 5}{(x - 2)(x + 1)} = \frac{\frac{11}{3}}{x - 2} + \frac{-\frac{2}{3}}{x + 1} \]
Summary
Partial fraction decomposition is a powerful tool in mathematics, especially for integration. By breaking down rational functions into more manageable pieces, you can simplify complex expressions and solve equations more effectively. Understanding how to decompose rational functions into partial fractions, recognizing distinct linear, repeated linear, distinct quadratic, and repeated quadratic special cases plays a crucial role in mastering this technique.
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