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: 13

Answer

$$6 - 3\ln 4$$

Work Step by Step

$$\eqalign{ & \int_0^3 {\frac{{2x - 1}}{{x + 1}}} dx \cr & {\text{by the long division}} \cr & = \int_0^3 {\left( {2 - \frac{3}{{x + 1}}} \right)} dx \cr & {\text{integrate}} \cr & = \left. {\left( {2x - 3\ln \left| {x + 1} \right|} \right)} \right|_0^3 \cr & {\text{using the fundamental theorem}} \cr & = \left( {2\left( 3 \right) - 3\ln \left| {3 + 1} \right|} \right) - \left( {2\left( 0 \right) - 3\ln \left| {0 + 1} \right|} \right) \cr & {\text{simplify}} \cr & = \left( {6 - 3\ln \left| 4 \right|} \right) - \left( {0 - 3\ln \left| 1 \right|} \right) \cr & = 6 - 3\ln 4 \cr} $$
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