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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![**Problem: Integrate the function.**
\[ \int \frac{\sqrt{x^2 - 16}}{x} \, dx \]
**Options:**
- **A.** \[ \left[ \frac{\sqrt{x^2 - 16}}{16} - \sec^{-1}\left(\frac{x}{4}\right) \right] + C \]
- **B.** \[ 4 \ln \left| \sqrt{x^2 - 16} - \left(\frac{x}{4}\right) \right| + C \]
- **C.** \[ 4 \left[ \frac{\sqrt{x^2 - 16}}{4} - \sec^{-1}\left(\frac{x}{4}\right) \right] + C \]
- **D.** \[ 4 \left[ \frac{\sqrt{x^2 - 16}}{4} - \sin^{-1}\left(\frac{x}{4}\right) \right] + C \]
**Explanation of options:**
Each option provides a different solution form involving integrals with inverse trigonometric functions and logarithms, where \( C \) represents the constant of integration. Options C and D use the expression in brackets with inverse secant or inverse sine functions, respectively.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F43abe65f-7e5f-4ca7-a91c-ab097f3acb86%2F523cacd4-dfef-4195-807e-f0bd696dc198%2Fqxhnqie_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem: Integrate the function.**
\[ \int \frac{\sqrt{x^2 - 16}}{x} \, dx \]
**Options:**
- **A.** \[ \left[ \frac{\sqrt{x^2 - 16}}{16} - \sec^{-1}\left(\frac{x}{4}\right) \right] + C \]
- **B.** \[ 4 \ln \left| \sqrt{x^2 - 16} - \left(\frac{x}{4}\right) \right| + C \]
- **C.** \[ 4 \left[ \frac{\sqrt{x^2 - 16}}{4} - \sec^{-1}\left(\frac{x}{4}\right) \right] + C \]
- **D.** \[ 4 \left[ \frac{\sqrt{x^2 - 16}}{4} - \sin^{-1}\left(\frac{x}{4}\right) \right] + C \]
**Explanation of options:**
Each option provides a different solution form involving integrals with inverse trigonometric functions and logarithms, where \( C \) represents the constant of integration. Options C and D use the expression in brackets with inverse secant or inverse sine functions, respectively.
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