Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises: 21

Answer

$F(x) = x^2 + \frac{1}{x} + 4x + C$, $(x\gt 0)$

Work Step by Step

$F(x) = \int \frac{2x^4 + 4x^3 -x}{x^3}dx$ $F(x) = \int \frac{x(2x^3 + 4x^2 -1)}{x^3}dx$ Simplify: $F(x) = \int \frac{2x^3 + 4x^2 -1}{x^2}dx$ $F(x) = \int (2x^3 + 4x^2 - 1) (x^{-2}) dx$ $F(x) = \int 2x^3x^{-2} + 4x^2x^{-2} - 1x^{-2} dx$ $F(x) = \int 2x + 4x^0 - 1x^{-2} dx$ $F(x) = \int 2x + 4(1) - 1x^{-2} dx$ $F(x) = \int 2x + 4 - x^{-2} dx$ $F(x) = \int 2(\frac{1}{2} x^2 + C) + \int -x^{-2}dx + \int 4dx$ $F(x) = \int 2(\frac{x^2}{2} + C) - \int x^{-2}dx + \int 4dx$ $F(x) = \int (x^2 + C) - (-x^{-1} + C) + (4x + C)$ $F(x) = x^2 + \frac{1}{x} + 4x + C$, $(x\gt 0)$
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