Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises: 22

Answer

$\frac 1 3 (3x^2+6x)^{1/2}+C$

Work Step by Step

$\int \frac {x+1}{\sqrt {3x^2+6x}}dx$ $u=3x^2+6x$ $\frac 1 6 du=x+1$ $\frac1 6 \int \frac{du}{u^{1/2}}$ $\frac 1 6 \times \frac{u^{1/2}}{1/2}$ $\frac 1 3 u^{1/2}+C$ $\frac 1 3 (3x^2+6x)^{1/2}+C$
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