Calculus: Early Transcendentals (2nd Edition)

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

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises - Page 238: 24

Answer

$$3{u^{ - 2}} - 4{u^2} + 1$$

Work Step by Step

$$\eqalign{ & \int {\left( {3{u^{ - 2}} - 4{u^2} + 1} \right)} du \cr & {\text{by the power rule for indefinite integrals}} \cr & {\text{ = }}\frac{{3{u^{ - 1}}}}{{ - 1}} - \frac{{4{u^3}}}{3} + u + C \cr & {\text{ = }} - 3{u^{ - 1}} - \frac{{4{u^3}}}{3} + u + C \cr & {\text{ = }} - \frac{3}{u} - \frac{{4{u^3}}}{3} + u + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{du}}\left( { - 3{u^{ - 1}} - \frac{{4{u^3}}}{3} + u + C} \right) \cr & {\text{ = }} - 3\left( { - 1} \right){u^{ - 2}} - \frac{{4\left( 3 \right){u^2}}}{3} + 1 + 0 \cr & {\text{simplify}} \cr & {\text{ = }}3{u^{ - 2}} - 4{u^2} + 1 \cr} $$
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