Calculus: Early Transcendentals (2nd Edition)

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

Chapter 6 - Applications of Integration - Review Exercises: 59

Answer

$$\frac{1}{2}\ln \left( {{x^2} + 8x + 25} \right) + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{x + 4}}{{{x^2} + 8x + 25}}} dx \cr & {\text{completing the square}} \cr & = \int {\frac{{x + 4}}{{{x^2} + 8x + 16 + 9}}} dx \cr & = \int {\frac{{x + 4}}{{{{\left( {x + 4} \right)}^2} + 9}}} dx \cr & {\text{substitute }}u = x + 4,{\text{ }}du = dx \cr & = \int {\frac{u}{{{u^2} + 9}}} du \cr & = \frac{1}{2}\int {\frac{{2u}}{{{u^2} + 9}}} du \cr & {\text{find the antiderivative}} \cr & = \frac{1}{2}\ln \left| {{u^2} + 9} \right| + C \cr & = \frac{1}{2}\ln \left( {{u^2} + 9} \right) + C \cr & {\text{replace back }}u = x + 4 \cr & = \frac{1}{2}\ln \left( {{{\left( {x + 4} \right)}^2} + 9} \right) + C \cr & = \frac{1}{2}\ln \left( {{x^2} + 8x + 25} \right) + C \cr} $$
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