Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.6 Other Integration Strategies - 7.6 Exercises: 73

Answer

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

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{{{\tan }^{ - 1}}\,x}}{{{x^2}}}\,dx} \hfill \\ \hfill \\ Integrate\,\,by\,\,tables,\,\,use\,\,the\,\,formulas \hfill \\ \hfill \\ \int_{}^{} {{x^n}{{\tan }^{ - 1}}xdx} = \frac{1}{{n + 1}}\,\,\left[ {{x^{n + 1}}{{\tan }^{ - 1}}x - \int_{}^{} {\frac{{{x^{n + 1}}}}{{1 + {x^2}}}\,dx} } \right] + C \hfill \\ \hfill \\ \,\,\int_{}^{} {\frac{1}{{x\,\left( {{x^2} + {a^2}} \right)}}dx} = \frac{1}{{2{a^2}}}\ln \left| {\frac{{{x^2}}}{{{a^2} + {x^2}}}} \right| + C \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {{x^{ - 2}}{{\tan }^{ - 1}}xdx} = \,\,\left[ {\frac{1}{x}{{\tan }^{ - 1}}x - \int_{}^{} {\frac{1}{{x\,\left( {1 + {x^2}} \right)}}dx} } \right] \hfill \\ \hfill \\ \int_{}^{} {{x^{ - 2}}{{\tan }^{ - 1}}xdx} = - \frac{1}{x}{\tan ^{ - 1}}x + \frac{1}{2}\ln \left| {\frac{{{x^2}}}{{1 + {x^2}}}} \right| + C \hfill \\ \hfill \\ \,\,using\,\,a\,\,\,CAS\,\,we\,\,get\,\,the\,\,same\,\,result \hfill \\ \hfill \\ \int_{}^{} {{x^{ - 2}}{{\tan }^{ - 1}}xdx} = - \frac{1}{x}{\tan ^{ - 1}}x + \frac{1}{2}\ln \left| {\frac{{{x^2}}}{{1 + {x^2}}}} \right| + C \hfill \\ \hfill \\ \end{gathered} \]
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