54x + 21 Evaluate dx where 9x2 + 7x + 8 > 0. | 9x² + 7x + 8 First substitute u = 54x + 21 -dx 7x + 8 Then du 9x2 + Now integrate with respect to u to get + C 54x + 21 -dx So 9x2 + 7x + 8 + C %3D
54x + 21 Evaluate dx where 9x2 + 7x + 8 > 0. | 9x² + 7x + 8 First substitute u = 54x + 21 -dx 7x + 8 Then du 9x2 + Now integrate with respect to u to get + C 54x + 21 -dx So 9x2 + 7x + 8 + C %3D
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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Question
![Evaluate
\[
\int \frac{54x + 21}{9x^2 + 7x + 8} \, dx
\]
where \(9x^2 + 7x + 8 > 0\).
First substitute \( u = \, \underline{\qquad\qquad\qquad\qquad\qquad} \)
Then
\[
\int \frac{54x + 21}{9x^2 + 7x + 8} \, dx = \int \, \underline{\qquad\qquad\qquad\qquad} \, du
\]
Now integrate with respect to \( u \) to get
\[
\underline{\qquad\qquad\qquad\qquad\qquad\qquad} + C
\]
So
\[
\int \frac{54x + 21}{9x^2 + 7x + 8} \, dx = \underline{\qquad\qquad\qquad\qquad\qquad\qquad} + C
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe8f98c03-2dea-45fe-9e7d-c001811b98bd%2Feb64cf54-dc44-4331-9da3-a6cd724b417b%2Fkxyrzao_processed.png&w=3840&q=75)
Transcribed Image Text:Evaluate
\[
\int \frac{54x + 21}{9x^2 + 7x + 8} \, dx
\]
where \(9x^2 + 7x + 8 > 0\).
First substitute \( u = \, \underline{\qquad\qquad\qquad\qquad\qquad} \)
Then
\[
\int \frac{54x + 21}{9x^2 + 7x + 8} \, dx = \int \, \underline{\qquad\qquad\qquad\qquad} \, du
\]
Now integrate with respect to \( u \) to get
\[
\underline{\qquad\qquad\qquad\qquad\qquad\qquad} + C
\]
So
\[
\int \frac{54x + 21}{9x^2 + 7x + 8} \, dx = \underline{\qquad\qquad\qquad\qquad\qquad\qquad} + C
\]
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