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Next, use the substitution w = x² – 9 to find the integral: SZ, da = S 1/w dw = In(w) +C = In(x**2-9) +C. (For the second answer blank, give your antiderivative in terms of the variable w. Again, note that you should not include the +C in your answer.)

Next, use the substitution w = x² – 9 to find the integral: SZ, da = S 1/w dw = In(w) +C = In(x**2-9) +C. (For the second answer blank, give your antiderivative in terms of the variable w. Again, note that you should not include the +C in your answer.)

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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Next, use the substitution w = x2 9 to find the integral: S, da = S 1/w dw = In(w) +C = In(x**2-9) +C. (For the second answer blank, give your antiderivative in terms of the variable w. Again, note that you should not include the +C in your answer.)
Transcribed Image Text:Next, use the substitution w = x2 9 to find the integral: S, da = S 1/w dw = In(w) +C = In(x**2-9) +C. (For the second answer blank, give your antiderivative in terms of the variable w. Again, note that you should not include the +C in your answer.)
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