Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 4 - Integration - 4.1 Exercises: 24

Answer

$= \frac{16}{5}t^{5} + 8t^{3} + 9t + C$

Work Step by Step

$\int (4t^{2}+3)^{2}dt$ $\int (4t^{2}(4t^{2} + 3) + 3(4t^{2} + 3)) dt$ $\int (16t^{4} + 12t^{2} + 12t^{2} + 9)dt$ $\int (16t^{4} + 24t^{2} + 9) dt$ $= \frac{16t^{5}}{5} + \frac{24t^{3}}{3} + 9t + C$ $= \frac{16}{5}t^{5} + 8t^{3} + 9t + C$
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