Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.6 Strategies for Integration - Exercises - Page 431: 34


$$\frac{(x+3) \sqrt{x^{2}+6 x}}{2}-\frac{9}{2} \ln \left|\frac{(x+3)+\sqrt{x^{2}+6 x}}{3}\right|$$

Work Step by Step

Given $$\int \sqrt{x^{2}+6 x} d x$$ Since $$\int \sqrt{x^{2}+6 x} d x =\int \sqrt{(x+3)^{2}-9} d x$$ Let $$ x+3=3\sec u,\ \ \ \ \ \ \ dx=3\sec u\tan udu$$ Then \begin{align*} \int \sqrt{x^{2}+6 x} d x&=3\int \sqrt{9\sec^2u-9}\sec u\tan udu \\ &=9\int \tan^2u \sec udu\\ &=9\int( \sec^3 u-\sec u )du,\ \\ & \text{From the table: }\\ &=9\left(\frac{1}{2}\sec \left(u\right)\tan \left(u\right)-\frac{1}{2}\ln \left|\tan \left(u\right)+\sec \left(u\right)\right|+C\right)\\ &= \frac{(x+3) \sqrt{x^{2}+6 x}}{2}-\frac{9}{2} \ln \left|\frac{(x+3)+\sqrt{x^{2}+6 x}}{3}\right| \end{align*}
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