University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Practice Exercises - Page 279: 114


$ \dfrac{-5}{x}+2 \tan^{-1} x+c$

Work Step by Step

Calculate the anti-derivative. Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ and $ \int \dfrac{1}{1+x^2} dx=tan^{-1} x+c$ where $c$ is a constant of proportionality. Then $\int (\dfrac{5}{x^2}+\dfrac{2}{x^2+1}) dx=5 \int x^{-2} dx+2 \tan^{-1} x +c$ or, $= \dfrac{5x^{-2+1}}{-2+1}+2 \tan^{-1} x+c$ Thus, $= \dfrac{-5}{x}+2 \tan^{-1} x+c$
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