Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.4 Derivatives of Logarithmic Functions - 6.4 Exercises - Page 438: 74


$\int_4^9(\sqrt x+{\frac{1}{\sqrt x}})^{2}dx=\frac{85}{2}+ln\frac{9}{4}$

Work Step by Step

Evaluate the integral $\int_4^9(\sqrt x+{\frac{1}{\sqrt x}})^{2}dx$. $=\int_4^9(x+\frac{1}{x}+2)dx$ $=\int_4^9(x)dx+\int_4^9(\frac{1}{x})dx+2 \int_4^9(1)dx$ $=[\frac{x^{2}}{2}]_4^9+ln[x]_4^9+2 [x]_4^9$ $=\frac{1}{2}[9^{2}-4^{2}]+[ln9-ln4]+2 [9-4]$ Hence, $\int_4^9(\sqrt x+{\frac{1}{\sqrt x}})^{2}dx=\frac{85}{2}+ln\frac{9}{4}$
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