Calculus 10th Edition

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

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises: 6

Answer

$\int \frac{du}{u}$ $u= t^{2}+t-4$

Work Step by Step

Try it: $du=(2t+1)dt$ $\int \frac{du}{u}=\ln |u| +C$ $\ln |t^{2}+t-4| +C$ $\int \frac{du}{u}$ $u= t^{2}+t-4$
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