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Fractional powers Use the indicated substitution to

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

Solution for problem 63E 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 63E

Fractional powers Use the indicated substitution to convert the given integral to an integral of a rational function. Evaluate the resulting integral.

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Problem 63EFractional powers Use the indicated substitution to convert the given integral to an integral of a rational function. Evaluate the resulting integral. Answer;Step-1; Definition of a Rational Function; A rational function is a function that is a fraction and has a property that both its numerator and denominator are polynomials . In other words , R(x) is a rational function if R(x) = where p(x) and q(x) are both polynomials , and q(x) recall that a polynomial is any function of the form f(x) = a +bx+ c+.............. +n, where a,b , c ……………….n are all real numbers and the exponents of each x is a non -negative integer. Proper 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; Improper fraction definition; In a rational fraction , if the degree of f(x) the degree of g(x) , then the rational fraction is called an improper fraction. If an improper rational fraction is given for splitting into partial fractions , we first divide f(x) with g(x) till we obtain a remainder R(x) of lower degree than g(x). First we express the fraction in the form = quotient + Then we resolve the final proper fraction into partial fractions. Step-2; The given integral is ; dx, x = ………….(1) In the given integral , integrand is a rational function. Now , we have to evaluate dx For our convenience take substitution method. Put x = , then = = , = and differentiation of x = is (x) = 1 = 6 dx = 6 du………………..(2) From (1) and (2) , the given integral becomes; dx = 6 du , since from(2). = 6 du………….(3) , since = = Step-3; From(3) , is an improper fraction. First we express the fraction in the form = quotient +. Then we resolve the final proper fraction into partial fractions , since from step -1. Therefore , = = - = - , since + =( - ……………(4) From(3) , (4) ; 6 du = 6(( - )du = 6[ du] = 6[ - +u - ln(u+1)]+C, since = +c , = ln(|x|) = 6[ - + - ln(+1)] +c , since from(2). = 2- 3+ 6 - 6ln(+1)] +c Therefore , from(1) and(3); dx = 6 du = 2- 3+ 6 - 6ln(+1)] +c Therefore , dx = 2- 3+ 6 - 6ln(+1)] +c

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Chapter 7.4, Problem 63E is Solved
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Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

The full step-by-step solution to problem: 63E from chapter: 7.4 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. Since the solution to 63E from 7.4 chapter was answered, more than 245 students have viewed the full step-by-step answer. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. This full solution covers the following key subjects: integral, convert, Fractional, function, given. This expansive textbook survival guide covers 112 chapters, and 5248 solutions. The answer to “Fractional powers Use the indicated substitution to convert the given integral to an integral of a rational function. Evaluate the resulting integral.” is broken down into a number of easy to follow steps, and 22 words. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1.

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Fractional powers Use the indicated substitution to