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: 32

Answer

\[\,\frac{1}{2}\,\left( {\ln x\sqrt {{{\ln }^2}x + 4} + 4\ln \left| {\ln x + \sqrt {{{\ln }^2}x + 4} } \right|} \right) + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{\sqrt {{{\ln }^2}x + 4} }}{x}} dx \hfill \\ \hfill \\ set\,\,u = \ln x\,\,\, \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \int_{}^{} {\frac{{\sqrt {{{\ln }^2}x + 4} }}{x}} dx = \int_{}^{} {\sqrt {{u^2} + {2^2}} du} \hfill \\ \hfill \\ use\,\,\,the\,\,formula \hfill \\ \hfill \\ \,\int_{}^{} {\sqrt {{u^2} + {a^2}} \,du = \frac{1}{2}\,\left( {u\sqrt {{u^2} + {a^2}} + {a^2}\ln \left| {u + \sqrt {{u^2} + {a^2}} } \right|} \right)} + C \hfill \\ \hfill \\ \,u = \ln x\,,\,\,\,therefore \hfill \\ \hfill \\ \int_{}^{} {\frac{{\sqrt {{{\ln }^2}x + 4} }}{x}} dx = \,\frac{1}{2}\,\left( {\ln x\sqrt {{{\ln }^2}x + 4} + 4\ln \left| {\ln x + \sqrt {{{\ln }^2}x + 4} } \right|} \right) + C \hfill \\ \hfill \\ \end{gathered} \]
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