Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.4 Trigonometric Substitutions - 7.4 Exercises: 36

Answer

$$\frac{{{y^3}}}{3} - y + {\tan ^{ - 1}}y + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{{y^4}}}{{{y^2} + 1}}dy} \cr & {\text{expanding integrand}} \cr & = \int {\left( {{y^2} - 1 + \frac{1}{{{y^2} + 1}}} \right)dy} \cr & {\text{sum rule}} \cr & = \int {{y^2}dy} - \int {dy} + \int {\frac{1}{{{y^2} + 1}}dy} \cr & = \int {{y^2}dy} - \int {dy} + \int {\frac{1}{{{y^2} + 1}}dy} \cr & {\text{integrating}} \cr & {\text{use power rule}} \cr & {\text{use }}\int {\frac{{du}}{{{u^2} + {a^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\frac{u}{a} + c \cr & = \frac{{{y^3}}}{3} - y + {\tan ^{ - 1}}y + C \cr} $$
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