Answer
$$-\frac{\pi}{6}$$
Work Step by Step
We have
$$\int^{-2/ \sqrt 3}_{-2}\frac{ dx}{|x|\sqrt{x^2-1}}=\sec^{-1}x|^{-2/ \sqrt 3}_{-2}\\
=\sec^{-1}\frac{-2}{\sqrt 3}-\sec^{-1}(-2)=\frac{5\pi}{6}-\frac{2\pi}{3}=-\frac{\pi}{6}.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 56
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