Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Explain why $\int \frac{2 x}{x^{2} + 1} d x$ and $\int \frac{1}{x \ln x} d x$ are essentially the same integral after a change of variables.

Both integrals turn into $\int \frac{1}{u} d u$ after a change of variables; $u = x^{2} + 1$ in the first case, $u = \ln x$ in the second.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information

Explain why $\int \frac{2 x}{x^{2} + 1} d x$ and role="math" localid="1648740843374" $\int \frac{1}{x \ln x} d x$ are essentially the same integral after a change of variables.

Step 2. Firstly changing the variable of ∫2xx2+1dx.

Let $x^{2} + 1 = u$

$u = x^{2} + 1 \frac{d u}{d x} = 2 x d u = 2 x d x$

This substitution changes the integral into

role="math" localid="1648740683170" $\int \frac{2 x}{x^{2} + 1} d x = \int \frac{1}{u} d u$

Step 3. Now changing the integral ∫1xlnxdx.

Let $\ln x = u$

$u = \ln x \frac{d u}{d x} = \frac{1}{x} d u = \frac{1}{x} d x$

This substitution changes the integral into

localid="1648740902640" $\int \frac{1}{x \ln x} d x = \int \frac{1}{u} d u$

Both integrals turn into $\int \frac{1}{u} d u$ after a change of variables; $u = x^{2} + 1$ in the first case, $u = \ln x$ in the second.

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Calculus

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Short Answer

Explain why $\int \frac{2 x}{x^{2} + 1} d x$ and $\int \frac{1}{x \ln x} d x$ are essentially the same integral after a change of variables.

Step by Step Solution

Step 1. Given Information

Step 2. Firstly changing the variable of ∫2xx2+1dx.

Step 3. Now changing the integral ∫1xlnxdx.

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Short Answer

Explain why ∫2xx2+1dx and ∫1xlnxdx are essentially the same integral after a change of variables.

Step by Step Solution

Step 1. Given Information

Step 2. Firstly changing the variable of ∫2xx2+1dx.

Step 3. Now changing the integral ∫1xlnxdx.

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Explain why $\int \frac{2 x}{x^{2} + 1} d x$ and $\int \frac{1}{x \ln x} d x$ are essentially the same integral after a change of variables.