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# Simple linear factors Evaluate the following integrals. ISBN: 9780321570567 2

## Solution for problem 9E Chapter 7.4

Calculus: Early Transcendentals | 1st Edition

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Problem 9E

Simple linear factors Evaluate the following integrals.

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Problem 9ESimple linear factors Evaluate the following integrals. Answer;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-2 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 non- repeated linear factors. Thus , from the above step the given fraction can be written as; = + = Thus , = …………..(2) Equating the numerator of L.H.S to the numerator of R.H.S Then , ( 1) = A(x+2) +B(x-1) 1 = (A+B)x+ (2A-B) A+B = 0 , and (2A-B) = 1.So , A = - B Solving A+B=1, 2A- B = 0 , then A = , and A+B= 0 A = 0 -B = 0 -= - B = ………..(3) Therefore, from(2) and(3) = = + , since from(1) = [-] Therefore , = [-] From(1) ; dx = [-] dx = [ dx - dx] = [ ln(|x-1|) - ln(|x+2|)]+C, since dx = ln(x) = ln(| )+C , since ln(a) -ln(b) = ln(a/b) Therefore , dx = ln(|| )+C

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##### ISBN: 9780321570567

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Simple linear factors Evaluate the following integrals.