University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.4 - Integration of Rational Functions by Partial Fractions - Exercises - Page 446: 48



Work Step by Step

$$I=\int\frac{1}{x\sqrt{x+9}}dx$$ We set $u^2=x+9$, so $$2udu=dx$$ Also, $x=u^2-9$ and $\sqrt{x+9}=u$ $$I=\int\frac{2udu}{(u^2-9)u}=\int\frac{2}{u^2-9}du$$ $$I=\int\frac{2}{(u-3)(u+3)}du$$ $$I=\frac{1}{3}\int\frac{6}{(u-3)(u+3)}du$$ $$I=\frac{1}{3}\int\frac{(u+3)-(u-3)}{(u-3)(u+3)}du$$ $$I=\frac{1}{3}\Big(\int\frac{du}{u-3}-\int\frac{du}{u+3}\Big)$$ $$I=\frac{1}{3}(\ln|u-3|-\ln|u+3|)+C$$ $$I=\frac{1}{3}\ln|\frac{u-3}{u+3}|+C$$ $$I=\frac{1}{3}\ln|\frac{\sqrt{x+9}-3}{\sqrt{x+9}+3}|+C$$
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