Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.2 Integration by Parts - 7.2 Exercises: 19

Answer

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

Work Step by Step

\[\begin{gathered} \int_{}^{} {{{\tan }^{ - 1}}xdx} \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ v = x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,dv = dx \hfill \\ u = {\tan ^{ - 1}}x\,\, \to \,\,\,\,\,du = \frac{1}{{1 + {x^2}}} \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ = x{\tan ^{ - 1}}x - \int_{}^{} {\frac{x}{{1 + {x^2}}}} dx \hfill \\ \hfill \\ rewrite \hfill \\ \hfill \\ x{\tan ^{ - 1}}x - \frac{1}{2}\int_{}^{} {\frac{{2x}}{{1 + {x^2}}}dx} \hfill \\ \hfill \\ integrate\,\,{\text{using }}\int {\frac{{du}}{u} = \ln u + C} \hfill \\ \hfill \\ = x{\tan ^{ - 1}}x - \frac{1}{2}\ln \,\left( {1 + {x^2}} \right) + C \hfill \\ \hfill \\ \end{gathered} \]
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