Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 362: 26

Answer

$F\left( x \right) = 3\left( {{{\tan }^{ - 1}}x} \right) + 2x + C$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \frac{{2{x^2} + 5}}{{{x^2} + 1}} \cr & {\text{Using the long division we obtain }}\frac{{2{x^2} + 5}}{{{x^2} + 1}} = \frac{3}{{{x^2} + 1}} + 2,{\text{ so}} \cr & f\left( x \right) = \frac{3}{{{x^2} + 1}} + 2 \cr & f\left( x \right) = 3\left( {\frac{1}{{{x^2} + 1}}} \right) + 2 \cr & {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right){\text{ }} \cr & {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr & {\text{Function: }}\frac{1}{{1 + {x^2}}} \to {\text{Particular antiderivative: ta}}{{\text{n}}^{ - 1}}x \cr & {\text{Function: }}f\left( x \right) + h\left( x \right) \to {\text{Particular antiderivative: }}f\left( x \right) + h\left( x \right) \cr & {\text{And applying the Theorem 1}},{\text{ we obtain}} \cr & f\left( x \right) = 3\left( {\frac{1}{{{x^2} + 1}}} \right) + 2 \cr & F\left( x \right) = 3\left( {{{\tan }^{ - 1}}x} \right) + 2x + C \cr} $$
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