Calculus: Early Transcendentals (2nd Edition)

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

Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 20

Answer

$$\frac{{{{\ln }^5}2}}{{10}}$$

Work Step by Step

$$\eqalign{ & \int_0^1 {\frac{{y{{\ln }^4}\left( {{y^2} + 1} \right)}}{{{y^2} + 1}}} dy \cr & {\text{substitute }}u = \ln \left( {{y^2} + 1} \right),{\text{ }}du = \frac{{2y}}{{{y^2} + 1}}dx \cr & {\text{express the limits in terms of }}u \cr & y = 0{\text{ implies }}u = \ln \left( {{0^2} + 1} \right) = 0 \cr & y = 1{\text{ implies }}u = \ln \left( {{1^2} + 1} \right) = \ln 2 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_0^1 {\frac{{y{{\ln }^4}\left( {{y^2} + 1} \right)}}{{{y^2} + 1}}} dy = \frac{1}{2}\int_0^{\ln 2} {{u^4}du} \cr & {\text{find the antiderivative}} \cr & = \left. {\frac{1}{2}\left( {\frac{{{u^5}}}{5}} \right)} \right|_0^{\ln 2} \cr & = \frac{1}{{10}}\left. {\left( {{u^5}} \right)} \right|_0^{\ln 2} \cr & {\text{use the fundamental theorem}} \cr & = \frac{1}{{10}}\left( {{{\ln }^5}2 - {0^5}} \right) \cr & = \frac{{{{\ln }^5}2}}{{10}} \cr} $$
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