Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 32

Answer

$$2\ln \left| {2{x^2} + 3x} \right| + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{8x + 6}}{{2{x^2} + 3x}}} dx \cr & {\text{substitute }}u = 2{x^2} + 3x,{\text{ }}du = \left( {4x + 3} \right)dx{\text{ }} \cr & {\text{ }}dx = \frac{{du}}{{4x + 3}} \cr & \int {\frac{{8x + 6}}{{2{x^2} + 3x}}} dx = \int {\frac{{8x + 6}}{u}} \frac{{du}}{{4x + 3}} \cr & = \int {\frac{{2\left( {4x + 3} \right)}}{u}} \frac{{du}}{{4x + 3}} \cr & = 2\int {\frac{{du}}{u}} \cr & {\text{find the antiderivative}} \cr & = 2\ln \left| u \right| + C \cr & {\text{ with}}\,\,\,u = 2{x^2} + 3x \cr & = 2\ln \left| {2{x^2} + 3x} \right| + C \cr} $$
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