Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Review - Exercises - Page 423: 27


$\int \frac{x+2}{\sqrt{x^2+4x}}~dx = \sqrt{x^2+4x}+C$

Work Step by Step

$\int \frac{x+2}{\sqrt{x^2+4x}}~dx$ Let $u = x^2+4x$ $\frac{du}{dx} = 2x+4$ $dx = \frac{du}{2x+4}$ $\int \frac{x+2}{\sqrt{u}}~\frac{du}{2x+4}$ $=\int \frac{1}{2}~\frac{du}{\sqrt{u}}$ $= \sqrt{u}+C$ $= \sqrt{x^2+4x}+C$
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