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> Read over Example 4 carefully. Then show how you decompose the following rational expression into its 2 partial fractions. HINT: You should be setting up and solving a system of two equations. 4x 2 2 + (x-1)(x+1) x-1 x+1 Find ¹S₁ = 4xdx (x-1)(x+1)

> Read over Example 4 carefully. Then show how you decompose the following rational expression into its 2 partial fractions. HINT: You should be setting up and solving a system of two equations. 4x 2 2 + (x-1)(x+1) x-1 x+1 Find ¹S₁ = 4xdx (x-1)(x+1)

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Find 1₁ 52-4 22² +-1 Solution Notice that the denominator can be factored as a product of linear factors: or 5ce-4 2²+2-1 In a case like this, where the numerator has a smaller degree than the denominator, we can write the given rational function as a sum of partial fractions: BÁ (+1)(2x - 1) Note 52-4 (x+1)(2x − 1) A 2+1 where A and B are constants. To find the values of A and B we multiply both sides of this equation by (x + 1)(2x - 1), obtaining + B 2z-1 5x-4 A(2-1) + B(z+1) 52-4-(2A+B) + (-A+B) The coefficients of a must be equal and the constant terms are also equal. So 561 (x+1)(2x - 1) 2A+B=5 and -A+B=-4 Solving this system of linear equations for A and B, we get A = 3 and B = -1, so 3 1 2+1 2x-1 Verify that this equation is correct by taking the fractions on the right side to a common denominator. Each of the resulting partial fractions is easy to integrate (using the substitutions w=* + 1 and u=27-1, respectively). So we have 3 /254-7-1dz = (2+1-22-1) dz
Transcribed Image Text:Find 1₁ 52-4 22² +-1 Solution Notice that the denominator can be factored as a product of linear factors: or 5ce-4 2²+2-1 In a case like this, where the numerator has a smaller degree than the denominator, we can write the given rational function as a sum of partial fractions: BÁ (+1)(2x - 1) Note 52-4 (x+1)(2x − 1) A 2+1 where A and B are constants. To find the values of A and B we multiply both sides of this equation by (x + 1)(2x - 1), obtaining + B 2z-1 5x-4 A(2-1) + B(z+1) 52-4-(2A+B) + (-A+B) The coefficients of a must be equal and the constant terms are also equal. So 561 (x+1)(2x - 1) 2A+B=5 and -A+B=-4 Solving this system of linear equations for A and B, we get A = 3 and B = -1, so 3 1 2+1 2x-1 Verify that this equation is correct by taking the fractions on the right side to a common denominator. Each of the resulting partial fractions is easy to integrate (using the substitutions w=* + 1 and u=27-1, respectively). So we have 3 /254-7-1dz = (2+1-22-1) dz
> Read over Example 4 carefully. Then show how you decompose the following rational expression into its 2 partial fractions. HINT: You should be setting up and solving a system of two equations. 4x 2 2 + (x-1)(x+1) x-1 x+1 Find ¹S₁ = 4xdx (x-1)(x+1)
Transcribed Image Text:> Read over Example 4 carefully. Then show how you decompose the following rational expression into its 2 partial fractions. HINT: You should be setting up and solving a system of two equations. 4x 2 2 + (x-1)(x+1) x-1 x+1 Find ¹S₁ = 4xdx (x-1)(x+1)
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