Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - Review - Exercises - Page 349: 24



Work Step by Step

Given $$\int \frac{x+2}{\sqrt{x^2+4x}}dx$$ Let $u=x^2+4x\ \Rightarrow \ du= 2(x+2)dx $ , then \begin{align*} \int \frac{x+2}{\sqrt{x^2+4x}}dx&=\frac{1}{2}\int \frac{du}{\sqrt{u}}\\ &=\sqrt{u}+c\\ &=\sqrt{x^2+4x}+c \end{align*}
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