Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.2 - Integration by Parts - Exercises 8.2 - Page 455: 51

Answer

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

Work Step by Step

$$\eqalign{ & \int x {\tan ^{ - 1}}xdx \cr & {\text{Using integration by parts method }} \cr & \,\,\,\,\,{\text{Let }}\,\,\,\,\,u = {\tan ^{ - 1}}x,\,\,\,\,du = \frac{1}{{1 + {x^2}}}\,\,\,dx \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = xdx,\,\,\,\,v = \frac{1}{2}{x^2} \cr & \cr & {\text{Integration by parts formula }}\int {udv} = uv - \int {vdu.{\text{ T}}} {\text{hen}}{\text{,}} \cr & \int x {\tan ^{ - 1}}xdx = \frac{1}{2}{x^2}{\tan ^{ - 1}}x - \int {\frac{1}{2}{x^2}} \left( {\frac{1}{{1 + {x^2}}}} \right)\,\,\,dx \cr & \int x {\tan ^{ - 1}}xdx = \frac{1}{2}{x^2}{\tan ^{ - 1}}x - \frac{1}{2}\int {\frac{{{x^2}}}{{1 + {x^2}}}} dx \cr & \int x {\tan ^{ - 1}}xdx = \frac{1}{2}{x^2}{\tan ^{ - 1}}x - \frac{1}{2}\int {\left( {\frac{{1 + {x^2}}}{{1 + {x^2}}} - \frac{1}{{1 + {x^2}}}} \right)} dx \cr & \int x {\tan ^{ - 1}}xdx = \frac{1}{2}{x^2}{\tan ^{ - 1}}x - \frac{1}{2}\int {\left( {1 - \frac{1}{{1 + {x^2}}}} \right)} dx \cr & \int x {\tan ^{ - 1}}xdx = \frac{1}{2}{x^2}{\tan ^{ - 1}}x - \frac{1}{2}\int {dx} + \frac{1}{2}\int {\frac{1}{{1 + {x^2}}}} dx \cr & {\text{integrating}} \cr & \int x {\tan ^{ - 1}}xdx = \frac{1}{2}{x^2}{\tan ^{ - 1}}x - \frac{1}{2}x + \frac{1}{2}{\tan ^{ - 1}}x + C \cr & {\text{factoring}} \cr & = \left( {\frac{1}{2}{x^2}{{\tan }^{ - 1}}x + \frac{1}{2}{{\tan }^{ - 1}}x} \right) - \frac{1}{2}x + C \cr & = \frac{1}{2}\left( {{x^2} + 1} \right){\tan ^{ - 1}}x - \frac{1}{2}x + C \cr} $$
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