Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.8 Improper Integrals - 7.8 Exercises: 23

Answer

\[ = \ln 2\]

Work Step by Step

\[\begin{gathered} \int_1^\infty {\frac{{dv}}{{v\,\left( {v + 1} \right)}}} \hfill \\ \hfill \\ integrating\,\,by\,\,partial\,\,fractions \hfill \\ \hfill \\ \int_{}^{} {\frac{{dv}}{{v\,\left( {v + 1} \right)}}} \,\, = \,\,\int_{}^{} {\,\left( {\frac{1}{v} - \frac{1}{{v + 1}}} \right)dv} \hfill \\ \hfill \\ = \ln \left| v \right| - \ln \left| {v + 1} \right| + C \hfill \\ \hfill \\ \log \,\,property \hfill \\ \hfill \\ = \ln \left| {\frac{v}{{v + 1}}} \right| + C \hfill \\ \hfill \\ use\,\,Definition\,\,of\,\,improper\,\,integral\,\, \hfill \\ \int_a^\infty {f\,\left( x \right)} \,dx = \,\,\,\,\,\mathop {\,\lim }\limits_{b \to \infty } \int_0^b {f\,\left( x \right)dx} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \int_1^\infty {\frac{{dv}}{{v\,\left( {v + 1} \right)}}} = \mathop {\lim }\limits_{b \to \infty } \int_a^b {\frac{{dv}}{{v\,\left( {v + 1} \right)}}} \hfill \\ \hfill \\ = \mathop {\lim }\limits_{b \to \infty } \,\,\ln \,\left| {\frac{v}{{v + 1}}} \right|_1^b \hfill \\ \hfill \\ use\,\,the\,\,ftc \hfill \\ \hfill \\ = \mathop {\lim }\limits_{b \to \infty } \,\left( {\ln \left| {\frac{b}{{b + 1}}} \right| - \ln \left| {\frac{1}{{1 + 1}}} \right|} \right) \hfill \\ \hfill \\ = \mathop {\lim }\limits_{b \to \infty } \,\left( {\ln \,\left( {1 - \frac{1}{{b + 1}}} \right) - \ln \frac{1}{2}} \right) \hfill \\ \hfill \\ {\text{Simplify}} \hfill \\ \hfill \\ = \,\left( {\ln \,\left( {1 - 0} \right) + \ln 2} \right) = \ln 2 \hfill \\ \end{gathered} \]
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