Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 66


$-(\tan t-1)^{-1}+c.$

Work Step by Step

Let $u=\tan t$, then $du=\sec^2t dt$. Now, we have, \begin{align*} \int\frac{\sec^2t }{(\tan t-1)^2} dt&=\int (u-1)^{-2} du\\ &=-(u-1)^{-1}+c\\ &=-(\tan t-1)^{-1}+c. \end{align*}
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