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: 56



Work Step by Step

We have \begin{align*} \int_{-3}^{-1} x(x^2+5)^{-2} dx&= \frac{1}{2}\int_{-3}^{-1} 2x(x^2+5)^{-2} dx\\ &=-\frac{1}{2} (x^2+5)^{-1}|_{-3}^{-1}\\ &=-\frac{1}{2}\frac{1}{6}+\frac{1}{2}\frac{1}{14}\\ &=-\frac{1}{21}. \end{align*}
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