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Repeated linear factors Evaluate the following | Ch 7.4 - 25E

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 25E Chapter 7.4

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 25E

Repeated linear factors Evaluate the following integrals.

Step-by-Step Solution:
Step 1 of 3

Problem 25EAnswer;Step-1Proper fraction definition ; In a rational fraction , if the degree of f(x) < the degree of g(x) , then the rational fraction is called a proper fraction. The sum of two proper fractions is a proper fraction. Example; Partial fractions Depending upon the nature of factors of Denominator ;1. When the denominator has non-repeated linear factors; A non - repeated linear factor (x-a) of denominator corresponds a partial fraction of the form , where A is a constant to be determined’ If g(x) = (x-a)(x-b)(x-c)............(x-n), then we assume that = ++ +...............+Where A, B, C,............N are constants which can be determined by equating the numerator of L.H.Sto the numerator of R.H.S , and substituting x = a,b ,c ….n.Step-21. When the denominator has repeated linear factors A repeated linear factor of denominator corresponds partial fractions of the form ; = + ++......................+Where A, B, C,............N are constants which can be determined by equating the numerator of L.H.Sto the numerator of R.H.S , and substituting x = a, we get N. Example; Step-3 The given integral is ; dx………………(1) Here the given integrand is of the form , and f(x) < g(x). Therefore , the given fraction is a proper fraction , and the denominator has repeated linear factors. Thus , from the above step the given fraction can be written as; = ++ = …………..(2) Equating the numerator of L.H.S to the numerator of R.H.S Then , x -5 = A+Ax +Bx+B +C = (A+C) +x(A+B) +B Therefore, A+C = 0 , A+B =1 and B = -5 , since equating the coefficients. Solving , A+C = 0 , A+B =1 and B = -5 then A= 1-B = 1-(-5) = 1+5 = 6 A+C =0 , then C = -A = -6 Thus , A = 6 , B = -5 and C =-6…………….(3) From(2) , (3) = ++ = ++ = --…………..(4) Therefore, from(1) and (4) dx = = dx -dx = 6dx -dx = 6ln(|x|) - - 6ln(|x+1|)+ C , since = +C , dx = ln(x)+c = 6 ln(||) - +c , ln(a) - ln(b) = ln(a/b) Therefore , dx = 6 ln(||) - +c

Step 2 of 3

Chapter 7.4, Problem 25E is Solved
Step 3 of 3

Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

This full solution covers the following key subjects: evaluate, factors, Integrals, Linear, repeated. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. Since the solution to 25E from 7.4 chapter was answered, more than 339 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 25E from chapter: 7.4 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. The answer to “Repeated linear factors Evaluate the following integrals.” is broken down into a number of easy to follow steps, and 7 words. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567.

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Repeated linear factors Evaluate the following | Ch 7.4 - 25E