Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.2* The Natural Logarithmic Functions - 6.2* Exercises: 69

Answer

$\int_1^e\frac{(x^{2}+x+1)}{x}dx=\frac{e^{2}-1+2e}{2}$

Work Step by Step

Evaluate the integral $\int_1^e\frac{(x^{2}+x+1)}{x}dx$. $\int_1^e(x+1+\frac{1}{x})dx$ $=\int_1^e(x)dx+\int_1^e(1)dx+\int_1^e(\frac{1}{x})dx$ $=[\frac{x^{2}}{2}]_1^e+[x]_1^e+ [lnx]_1^e$ $=\frac{1}{2}[e^{2}-1]+[e-1]+[lne-ln1]$ Hence, $\int_1^e\frac{(x^{2}+x+1)}{x}dx=\frac{e^{2}-1+2e}{2}$
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