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: 70



Work Step by Step

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