Answer
$$-\frac{2}{27}$$
Work Step by Step
We have
$$\int_{-4}^{- 2} 12x(x^2+2)^{-3}dx=
6\times\int_{-4}^{- 2} 2x(x^2+2)^{-3}dx\\
=\frac{6}{-2} (x^2+2)^{-2}|_{-4}^{-2}=-3(\frac{1}{6^2}-\frac{1}{18^2})=-\frac{2}{27}$$
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: 70
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