Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 7 - Principles Of Integral Evaluation - 7.6 Using Computer Algebra Systems And Tables Of Integrals - Exercises Set 7.6 - Page 531: 48

Answer

$$\frac{1}{2}\left( {2 + {x^2}} \right)\left[ {\ln \left( {2 + {x^2}} \right) - 1} \right] + C$$

Work Step by Step

$$\eqalign{ & \int {x\ln \left( {2 + {x^2}} \right)} dx \cr & {\text{Make an appropiate }}u{\text{ - substitution }} \cr & u = 2 + {x^2},\,\,\,\,\,\,\,du = 2xdx,\,\,\,\,\,\,dx = \frac{{du}}{{2x}} \cr & {\text{write in terms of }}u \cr & \int {x\ln \left( {2 + {x^2}} \right)} dx = \int {x\ln u} \left( {\frac{{du}}{{2x}}} \right) \cr & = \frac{1}{2}\int {\ln u} du \cr & {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr & {\text{By formula 11}} \cr & \left( {11} \right):\,\,\,\,\,\,\int {\ln u} du = u\ln u - u + C \cr & \frac{1}{2}\int {\ln u} du = \frac{1}{2}u\ln u - \frac{1}{2}u + C \cr & {\text{write in terms of }}x{\text{; replace }}2 + {x^2}{\text{ for }}u \cr & = \frac{1}{2}\left( {2 + {x^2}} \right)\ln \left( {2 + {x^2}} \right) - \frac{1}{2}\left( {2 + {x^2}} \right) + C \cr & {\text{factoring}} \cr & = \frac{1}{2}\left( {2 + {x^2}} \right)\left[ {\ln \left( {2 + {x^2}} \right) - 1} \right] + C \cr} $$
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