Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises: 25

Answer

$\int\frac{x^3 + 3x-4}{x^2} dx=\frac{1}{2}x^2 + 3ln|x| +\frac{4}{x}+C $

Work Step by Step

$\int\frac{x^3 + 3x-4}{x^2} dx= \int x + \frac{3}{x} -\frac{4}{x^2} dx = \int x + \frac{3}{x}-4x^{-2}dx = \frac{1}{2}x^2 + 3ln|x| - 4\times\frac{1}{-1}x^{-1}+C=\frac{1}{2}x^2 + 3ln|x| +\frac{4}{x}+C $
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