Answer
$-(\tan t-1)^{-1}+c.$
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*}
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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