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: 68

Answer

=1

Work Step by Step

\[\begin{gathered} \int_1^\infty {\frac{{\ln x}}{{{x^2}}}} \,dx \hfill \\ \hfill \\ use\,\,the\,\,Formula\,\,for\,\,in\,tegration\,\,by\,\,parts \hfill \\ \hfill \\ \int_{}^{} {udv = uv - \int_{}^{} {vdu} } \hfill \\ \hfill \\ set \hfill \\ du = \ln x\,\,\,\,\,\,\,then\,\,\,\,du = \frac{1}{x}dx \hfill \\ dv = \frac{1}{{{x^2}}}dx\,\,then\,\,\,\,v = - \frac{1}{x} \hfill \\ \hfill \\ integrating \hfill \\ \hfill \\ \int_{}^{} {\frac{{\ln x}}{{{x^2}}}} \,dx = \,\,\left[ { - \frac{1}{x}\ln x - \frac{1}{x}} \right] \hfill \\ \hfill \\ \int_1^\infty {\frac{{\ln x}}{{{x^2}}}} \,dx\,\, = \mathop {\lim }\limits_{a \to 0} \int_1^a {\frac{{\ln x}}{{{x^2}}}\,dx} \hfill \\ \hfill \\ = \mathop {\lim }\limits_{a \to 0} \,\,\left[ { - \frac{1}{x}\ln x - \frac{1}{x}} \right]_1^a \hfill \\ \hfill \\ use\,\,the\,\,ftc \hfill \\ \hfill \\ = \mathop {\lim }\limits_{a \to 0} \,\,\left[ { - \frac{1}{a}\ln a - \frac{1}{a} + 1} \right] \hfill \\ \hfill \\ evaluate\,\,the\,\,\,limit \hfill \\ \hfill \\ = 1 \hfill \\ \hfill \\ \end{gathered} \]
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