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: 7


$\ln |1+\sqrt x|+c$

Work Step by Step

Solve $\int\dfrac{dx}{2 \sqrt x+2x}$ or, $\int\dfrac{dx}{2 \sqrt x+2x}=\int\dfrac{dx}{2 \sqrt x(1+\sqrt x)}$ Let $p=1+\sqrt x$ and $dp=\dfrac{1}{2 \sqrt x }dx$ This implies $\int\dfrac{dx}{2 \sqrt x(1+\sqrt x)}=\int\dfrac{dp}{p}=\ln|p|+c$ Since, $p=1+\sqrt x$ Thus, $\int\dfrac{dx}{2 \sqrt x+2x}=\ln |1+\sqrt x|+c$
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