Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - Review Exercises - Page 594: 59

Answer

$$\frac{1}{8}\ln \left| {\frac{{x - 5}}{{x + 3}}} \right| + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{dx}}{{{x^2} - 2x - 15}}} \cr & {\text{completing the square}} \cr & \int {\frac{{dx}}{{{x^2} - 2x - 15}}} = \int {\frac{{dx}}{{{x^2} - 2x + 1 - 16}}} \cr & = \int {\frac{{dx}}{{\left( {{x^2} - 2x + 1} \right) - 16}}} \cr & = \int {\frac{{dx}}{{{{\left( {x - 1} \right)}^2} - 16}}} \cr & u = x - 1,{\text{ }}du = dx \cr & = \int {\frac{{du}}{{{u^2} - 16}}} \cr & = \int {\frac{{du}}{{\left( {u + 4} \right)\left( {u - 4} \right)}}} \cr & {\text{partial fractions}} \cr & \frac{1}{{\left( {u + 4} \right)\left( {u - 4} \right)}} = \frac{A}{{u + 4}} + \frac{B}{{u - 4}} \cr & 1 = A\left( {u - 4} \right) + B\left( {u + 4} \right) \cr & u = - 4 \to A = - \frac{1}{8} \cr & u = 4 \to B = \frac{1}{8} \cr & \frac{1}{{\left( {u + 4} \right)\left( {u - 4} \right)}} = - \frac{{1/8}}{{u + 4}} + \frac{{1/8}}{{u - 4}} \cr & = \int {\frac{{du}}{{\left( {u + 4} \right)\left( {u - 4} \right)}}} = \int {\left( { - \frac{{1/8}}{{u + 4}} + \frac{{1/8}}{{u - 4}}} \right)} du \cr & = \frac{1}{8}\ln \left| {u - 4} \right| - \frac{1}{8}\ln \left| {u + 4} \right| + C \cr & = \frac{1}{8}\ln \left| {\frac{{u - 4}}{{u + 4}}} \right| + C \cr & {\text{write in terms of }}x \cr & = \frac{1}{8}\ln \left| {\frac{{x - 1 - 4}}{{x - 1 + 4}}} \right| + C \cr & = \frac{1}{8}\ln \left| {\frac{{x - 5}}{{x + 3}}} \right| + C \cr} $$
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