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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How do i solve this step by step please. The answer should be 9/x + ln|x-9/x| +c
![The problem labeled as "8.5.31" involves an integral and its simplification process. The given integral is:
\[
\int \frac{81}{x^3 - 9x^2} \, dx
\]
This is simplified through a series of steps shown below:
1. Break down the denominator:
\[
\frac{81}{x^3 - 9x^2} = \frac{81}{x^2(x-9)}
\]
2. Perform partial fraction decomposition:
\[
\frac{81}{x^2(x-9)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-9}
\]
Here, \( A \), \( B \), and \( C \) are constants to be determined. This process involves expressing the original rational expression as a sum of simpler fractions.
The text on the image stops at setting up the partial fraction decomposition without solving for \( A \), \( B \), and \( C \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F25b167f9-3118-479d-a9fb-f8c446c41f1a%2F33ceb0cd-d203-430c-86d9-682b455ff1e5%2F92r283v_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The problem labeled as "8.5.31" involves an integral and its simplification process. The given integral is:
\[
\int \frac{81}{x^3 - 9x^2} \, dx
\]
This is simplified through a series of steps shown below:
1. Break down the denominator:
\[
\frac{81}{x^3 - 9x^2} = \frac{81}{x^2(x-9)}
\]
2. Perform partial fraction decomposition:
\[
\frac{81}{x^2(x-9)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-9}
\]
Here, \( A \), \( B \), and \( C \) are constants to be determined. This process involves expressing the original rational expression as a sum of simpler fractions.
The text on the image stops at setting up the partial fraction decomposition without solving for \( A \), \( B \), and \( C \).
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