Answer
$\sqrt{2x}+C$
Work Step by Step
$\int\frac{1}{\sqrt{2x}}dx=\int\frac{x^{-\frac{1}{2}}}{\sqrt{2}}dx=\frac{2\sqrt{x}}{\sqrt{2}}+C=\sqrt{2x}+C$
CHECK:
$\frac{d}{dx}\sqrt{2x}=\frac{1}{2}(2x)^{-\frac{1}{2}}\times2=\frac{1}{\sqrt{2x}}$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 4 - Integration - 4.5 Exercises - Page 301: 25
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