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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 7.4 - Problem 61e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 7.4 - Problem 61e

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Fractional powers Use the indicated | Ch 7.4 - 61E

ISBN: 9780321570567 2

Solution for problem 61E Chapter 7.4

Calculus: Early Transcendentals | 1st Edition

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Problem 61E

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 61EFractional 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+2 = ………….(1) In the given integral , integrand is is a rational function. Now , we have to evaluate dx For our convenience take substitution method. Put x+2 = , then = = u , and differentiation of x+2 = is (x+2) = 1+ 0 = 4 , since = n 1 = 4 dx= 4 du ………………..(2) From (1) and (2) , the given integral becomes; dx = 4 du , since from(2). = 4 du………….(3) 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) ; 4 du = 4(( - )du = 4[ du] = 4[ - +u - ln(u+1)]+C, since = +c , = ln(|x|) = 4[ - + - ln(+1)] +c , since from(2). = 4 - + - 4ln(+1)+C. Therefore , from(1) and(3); dx = 4 du = 4 - + - 4ln(+1)+C. Therefore , dx = 4 - + - 4ln(+1)+C.

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