Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 49

Answer

$$\frac{1}{2}\ln \left( {17} \right)$$

Work Step by Step

$$\eqalign{ & \int_0^4 {\frac{x}{{{x^2} + 1}}dx} \cr & {\text{substitute }}u = {x^2} + 1,{\text{ }}du = 2xdx \cr & {\text{express the limits in terms of }}u \cr & x = 0{\text{ implies }}u = {\left( 0 \right)^2} + 1 = 1 \cr & x = 4{\text{ implies }}u = {\left( 4 \right)^2} + 1 = 17 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_0^4 {\frac{x}{{{x^2} + 1}}dx} = \int_1^{16} {\frac{{1/2}}{u}du} \cr & {\text{find the antiderivative}} \cr & = \left. {\left( {\frac{1}{2}\ln \left| u \right|} \right)} \right|_1^{17} \cr & {\text{Use the fundamental theorem}} \cr & = \frac{1}{2}\ln \left| {17} \right| - \frac{1}{2}\ln \left| 1 \right| \cr & {\text{Simplify}} \cr & = \frac{1}{2}\ln \left( {17} \right) - \frac{1}{2}\ln \left( 1 \right) \cr & = \frac{1}{2}\ln \left( {17} \right) \cr} $$
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