University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 18


$\sqrt {\ln^2 x+1} +c$

Work Step by Step

Solve $\int \dfrac{\ln x dx}{x \sqrt {\ln^2 x+1}}$ Let us $p=\ln^2 x+1$ and $dp=2( \ln x)( \frac{1}{x}) dx$ This implies $\int \dfrac{\ln x dx}{x \sqrt {\ln^2 x+1}}=\frac{1}{2}\int \dfrac{1}{\sqrt p} dp $ $=p^{1/2} +c$ $=\sqrt p +c$ Hence,$\int \dfrac{\ln x dx}{x \sqrt {\ln^2 x+1}}=\sqrt {\ln^2 x+1} +c$
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