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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![### Integration Exercise: Finding Indefinite Integrals
#### Problem 14: Indefinite Integral Calculation
**Original Integral:**
\[ \int \frac{1}{(3x)^2} \, dx \]
**Steps to Solve:**
1. **Rewrite:**
Rewrite the integrand to a more convenient form for integration:
\[ \int \frac{1}{9x^2} \, dx \]
Since \((3x)^2 = 9x^2\).
2. **Integrate:**
Use the basic integral formula for \(\int x^{-n} dx = \frac{x^{-n+1}}{-n+1} + C\) for \(n \neq 1\):
\[ \int \frac{1}{9x^2} \, dx = \int \frac{1}{9} x^{-2} \, dx = \frac{1}{9} \int x^{-2} \, dx \]
\[ = \frac{1}{9} \left( \frac{x^{-2+1}}{-2+1} \right) + C = \frac{1}{9} \left( \frac{x^{-1}}{-1} \right) + C \]
3. **Simplify:**
Simplify the resulting expression:
\[ = \frac{1}{9} \left( -\frac{1}{x} \right) + C \]
\[ = -\frac{1}{9x} + C \]
**Answer:**
\[ \int \frac{1}{(3x)^2} \, dx = -\frac{1}{9x} + C \]
This exercise demonstrates the step-by-step process of manipulating and integrating a given function. Understanding these steps is crucial for effectively solving indefinite integrals.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F83b540d6-8dbe-47db-b28f-7c0bcbc5fc80%2Fc45f7da6-7e95-426c-b92f-2075a47354fe%2F1ikdadj.jpeg&w=3840&q=75)
Transcribed Image Text:### Integration Exercise: Finding Indefinite Integrals
#### Problem 14: Indefinite Integral Calculation
**Original Integral:**
\[ \int \frac{1}{(3x)^2} \, dx \]
**Steps to Solve:**
1. **Rewrite:**
Rewrite the integrand to a more convenient form for integration:
\[ \int \frac{1}{9x^2} \, dx \]
Since \((3x)^2 = 9x^2\).
2. **Integrate:**
Use the basic integral formula for \(\int x^{-n} dx = \frac{x^{-n+1}}{-n+1} + C\) for \(n \neq 1\):
\[ \int \frac{1}{9x^2} \, dx = \int \frac{1}{9} x^{-2} \, dx = \frac{1}{9} \int x^{-2} \, dx \]
\[ = \frac{1}{9} \left( \frac{x^{-2+1}}{-2+1} \right) + C = \frac{1}{9} \left( \frac{x^{-1}}{-1} \right) + C \]
3. **Simplify:**
Simplify the resulting expression:
\[ = \frac{1}{9} \left( -\frac{1}{x} \right) + C \]
\[ = -\frac{1}{9x} + C \]
**Answer:**
\[ \int \frac{1}{(3x)^2} \, dx = -\frac{1}{9x} + C \]
This exercise demonstrates the step-by-step process of manipulating and integrating a given function. Understanding these steps is crucial for effectively solving indefinite integrals.
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