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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 8.5.2

Expand the quotients in Exercises 1–8 by partial fractions.
(5x - 7) / (x² - 3x + 2)

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1
First, factor the denominator \(x^{2} - 3x + 2\) into its linear factors. To do this, find two numbers that multiply to \(2\) and add to \(-3\).
Rewrite the original expression as a sum of partial fractions with unknown coefficients: \(\frac{5x - 7}{(x - a)(x - b)} = \frac{A}{x - a} + \frac{B}{x - b}\), where \(a\) and \(b\) are the roots found from factoring.
Multiply both sides of the equation by the denominator \((x - a)(x - b)\) to clear the fractions, resulting in an equation involving polynomials: \(5x - 7 = A(x - b) + B(x - a)\).
Expand the right-hand side and collect like terms in \(x\). This will give you an equation of the form \(5x - 7 = (A + B)x - (Ab + Ba)\).
Equate the coefficients of corresponding powers of \(x\) on both sides to form a system of linear equations. Solve this system to find the values of \(A\) and \(B\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to express a rational function as a sum of simpler fractions with linear or quadratic denominators. This technique simplifies integration and other operations by breaking down complex fractions into manageable parts.
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Partial Fraction Decomposition: Distinct Linear Factors

Factoring Quadratic Denominators

Factoring the quadratic denominator into linear factors is essential for partial fraction decomposition. For example, x² - 3x + 2 factors into (x - 1)(x - 2), allowing the original fraction to be split into terms with these simpler denominators.
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Partial Fraction Decomposition: Irreducible Quadratic Factors

Setting Up and Solving Systems of Equations

After expressing the fraction as a sum of partial fractions with unknown coefficients, you multiply both sides by the denominator and equate coefficients of like terms. This process yields a system of linear equations to solve for the unknown constants.
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