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S √16 - x² dx X 4+16x2 A)√16-x2 - 4 In X C) √/16 - x² - sin-1 ] + c - C B)√16-x2+4 In 4+√16-x2 x2 √16-x2 D) -sin X + C + C

S √16 - x² dx X 4+16x2 A)√16-x2 - 4 In X C) √/16 - x² - sin-1 ] + c - C B)√16-x2+4 In 4+√16-x2 x2 √16-x2 D) -sin X + C + C

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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S 16-x2 X dx A) √16 - x2 - 4 In C)√16-x2-sin- 4+ 16 - x² X +C 0 B)√16- x2 + 4 In D) -sin- 4+√16-x2 x2 1/4)√16-x² X + C ܘܢ + C
Transcribed Image Text:S 16-x2 X dx A) √16 - x2 - 4 In C)√16-x2-sin- 4+ 16 - x² X +C 0 B)√16- x2 + 4 In D) -sin- 4+√16-x2 x2 1/4)√16-x² X + C ܘܢ + C
Evaluate using partial function ƒ =²+x²+x²-1-dx **—*—2 Α. ÷x² + 3x − ―ln |x + 2 + In |x-1| + c +x² + 5x − —ln ³x + 2| + †ln |x − 1] + c =x² + 5x − −ln |x + 2| + ―ln |x − 1| + c C. —-x² + 5x − −ln (x − 2| + --ln |x − 1| + c - + D 13 m
Transcribed Image Text:Evaluate using partial function ƒ =²+x²+x²-1-dx **—*—2 Α. ÷x² + 3x − ―ln |x + 2 + In |x-1| + c +x² + 5x − —ln ³x + 2| + †ln |x − 1] + c =x² + 5x − −ln |x + 2| + ―ln |x − 1| + c C. —-x² + 5x − −ln (x − 2| + --ln |x − 1| + c - + D 13 m
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