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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 7.4 - Problem 26e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 7.4 - Problem 26e

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# Setting up partial fraction decompositions Give the

ISBN: 9780321570567 2

## Solution for problem 26E Chapter 7.4

Calculus: Early Transcendentals | 1st Edition

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Problem 26E

Setting up partial fraction decompositions Give the appropriate form of the partial fraction decomposition for the following functions.

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Problem 26EAnswer;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 2) 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 , 2 = A-3Ax +Bx+C6 Cx+9C = (A+C) +x(-3A+B -6C) +9C Therefore, A+C = 0 , -3A+B-6C=0, and 9C = 2 , since equating the coefficients. Solving , A+C = 0 , -3A+B-6C =1 and C = then A = -C = -( ) B = 3A+6C = 3(- )+6( ) = = = Thus , A = - , B = and C = …………….(3) Therefore, from(2) , (3) = = ++ = (- ) + ()+ ………….(4) Therefore, from(1) and(4) dx = =dx + dx +dx = ln(|x-3|) ++ ln(|x|)+C , since = +C , dx = ln(x)+c = ln(||) - +C , since ln(a) - ln(b) = ln(a/b) Therefore, dx = ln(||) - +C

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##### ISBN: 9780321570567

The full step-by-step solution to problem: 26E from chapter: 7.4 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. The answer to “Setting up partial fraction decompositions Give the appropriate form of the partial fraction decomposition for the following functions.” is broken down into a number of easy to follow steps, and 18 words. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. This full solution covers the following key subjects: fraction, partial, appropriate, form, decompositions. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. Since the solution to 26E from 7.4 chapter was answered, more than 382 students have viewed the full step-by-step answer.

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