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c. The seemingly different answers are the same, because they differ by a constant. In fact, (x + 1)² 22 2 - - 2(x + 1) + ln x + 1| − 2 · − x + ln \x + 1) == 2

c. The seemingly different answers are the same, because they differ by a constant. In fact, (x + 1)² 22 2 - - 2(x + 1) + ln x + 1| − 2 · − x + ln \x + 1) == 2

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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I dont understand what they want me to do for part c

#TI a) 2 Sta X+1 • 71. Different methods Let I = 2 = $_+² du u = Scu-13² du 2 u S = Scu-1)(1-1) du (-۱ u S a. Evaluate / using the substitution u = x + 1. b. Evaluate / after first performing long division on the integrand. c. Reconcile the results in parts (a) and (b). dx u²²-u-u+1 du u 2 2 u u-2u+1 du u u S#1 x + 1 2u + 1 dx u-2 + 1 dx M = x + 1 x= μ-1 du = dx 2 = μ² - 2u + ln /u/+C 2 - dx. 2 = (x+1) - 2 (x+1) + ln (x+1)+c 2 s dx b) X 2 X+1 X+1 X-1 x² x + x - X -x+1 Seit I I dx de +- X+1 = Sxdx + S-idx + S ____ dx X+1 2 x -x + ln | X + 1 | + c 2 2 (X+ 1)² - 2 (X+1) + ln (x+1) - ( 22 −X + ln | X + ¹ \ ) 2
Transcribed Image Text:#TI a) 2 Sta X+1 • 71. Different methods Let I = 2 = $_+² du u = Scu-13² du 2 u S = Scu-1)(1-1) du (-۱ u S a. Evaluate / using the substitution u = x + 1. b. Evaluate / after first performing long division on the integrand. c. Reconcile the results in parts (a) and (b). dx u²²-u-u+1 du u 2 2 u u-2u+1 du u u S#1 x + 1 2u + 1 dx u-2 + 1 dx M = x + 1 x= μ-1 du = dx 2 = μ² - 2u + ln /u/+C 2 - dx. 2 = (x+1) - 2 (x+1) + ln (x+1)+c 2 s dx b) X 2 X+1 X+1 X-1 x² x + x - X -x+1 Seit I I dx de +- X+1 = Sxdx + S-idx + S ____ dx X+1 2 x -x + ln | X + 1 | + c 2 2 (X+ 1)² - 2 (X+1) + ln (x+1) - ( 22 −X + ln | X + ¹ \ ) 2
c. The seemingly different answers are the same, because they differ by a constant. In fact, (x + 1)² X 3 2(x + 1) + ln |x + 1| − (22²2 - x + ln x + 1| +11) 2 2 =
Transcribed Image Text:c. The seemingly different answers are the same, because they differ by a constant. In fact, (x + 1)² X 3 2(x + 1) + ln |x + 1| − (22²2 - x + ln x + 1| +11) 2 2 =
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