Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.1 Basic Approaches - 7.1 Exercises - Page 514: 23

Answer

$$\frac{1}{2}\ln \left( {{x^2} + 4} \right) + {\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{x + 2}}{{{x^2} + 4}}dx} \cr & {\text{split numerator}} \cr & = \int {\frac{x}{{{x^2} + 4}}dx} + \int {\frac{2}{{{x^2} + 4}}dx} \cr & = \frac{1}{2}\int {\frac{{2x}}{{{x^2} + 4}}dx} + 2\int {\frac{1}{{{x^2} + {{\left( 2 \right)}^2}}}dx} \cr & {\text{find the antiderivatives}}{\text{, }} \cr & {\text{formulas}}:{\text{ }}\int {\frac{{du}}{u}} = \ln \left| u \right| + C,{\text{ }}\int {\frac{{du}}{{{u^2} + {a^2}}} = \frac{1}{a}{{\tan }^{ - 1}}} \left( {\frac{u}{a}} \right) + C \cr & = \frac{1}{2}\ln \left| {{x^2} + 4} \right| + 2\left( {\frac{1}{2}{{\tan }^{ - 1}}\left( {\frac{x}{2}} \right)} \right) + C \cr & simplify \cr & = \frac{1}{2}\ln \left( {{x^2} + 4} \right) + {\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C \cr} $$
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