Explanation Integrate the given equation by rewriting it in a way that it can be integrated by parts to verify the equation. Given: Let I n = ∫ x n d x x 2 + 1 Calculation: I n = ∫ x n d x x 2 + 1 = ∫ x n − 2 ( x 2 + 1 − 1 ) d x x 2 + 1 To determine (b) Use the previous solution in (a) to calculate I n for 0 ≤ n ≤ 5 To determine (c) To show that, in general, I 2 n + 1 = x 2 n 2 n − x 2 n − 2 2 n − 2 + ... + ( − 1 ) n − 1 x 2 2 + ( − 1 ) n ln ( x 2 + 1 ) 2 + C I 2 n = x 2 n − 1 2 n − 1 − x 2 n − 3 2 n − 3 + ... + ( − 1 ) n − 1 x + ( − 1 ) n tan − 1 x + C

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Question

Chapter 8, Problem 67CRE

To determine

(a)

To prove that $I_{n} = \frac{x^{n - 1}}{n - 1} - I_{n - 2}$

To determine

(b)

Use the previous solution in (a) to calculate $I_{n}$ for $0 \leq n \leq 5$

To determine

(c)

To show that, in general, $I_{2 n + 1} = \frac{x^{2 n}}{2 n} - \frac{x^{2 n - 2}}{2 n - 2} + ... + {(- 1)}^{n - 1} \frac{x^{2}}{2} + {(- 1)}^{n} \frac{\ln (x^{2} + 1)}{2} + C$

$I_{2 n} = \frac{x^{2 n - 1}}{2 n - 1} - \frac{x^{2 n - 3}}{2 n - 3} + ... + {(- 1)}^{n - 1} x + {(- 1)}^{n} \tan^{- 1} x + C$

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Chapter 8, Problem 66CRE

Chapter 8, Problem 68CRE

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Chapter 8 Solutions

Calculus - Standalone book

Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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(a) To prove that I n = x n − 1 n − 1 − I n − 2

Solution Summary: The author explains how to calculate I n for 0

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(a)

To prove that $I_{n} = \frac{x^{n - 1}}{n - 1} - I_{n - 2}$

(b)

Use the previous solution in (a) to calculate $I_{n}$ for $0 \leq n \leq 5$

Want to see the full answer?

Chapter 8 Solutions

(a) To prove that I n = x n − 1 n − 1 − I n − 2

Solution Summary: The author explains how to calculate I n for 0

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(a) To prove that In=xn−1n−1−In−2

(b) Use the previous solution in (a) to calculate In for 0≤n≤5

(c) To show that, in general, I2n+1=x2n2n−x2n−22n−2+...+(−1)n−1x22+(−1)nln(x2+1)2+C I2n=x2n−12n−1−x2n−32n−3+...+(−1)n−1x+(−1)ntan−1x+C

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Chapter 8 Solutions

(a)

To prove that $I_{n} = \frac{x^{n - 1}}{n - 1} - I_{n - 2}$

(b)

Use the previous solution in (a) to calculate $I_{n}$ for $0 \leq n \leq 5$