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: 47



Work Step by Step

$$I=\int\frac{\sqrt{x+1}}{x}dx$$ We set $u^2=x+1$, so $$2udu=dx$$ Also, $x=u^2-1$ and $\sqrt{x+1}=u$ $$I=\int\frac{u}{u^2-1}(2udu)=\int\frac{2u^2}{u^2-1}du$$ $$I=\int\frac{2u^2-2}{u^2-1}du+\int\frac{2}{u^2-1}du$$ $$I=\int2du+\int\frac{2}{(u-1)(u+1)}du$$ $$I=2u+\int\frac{(u+1)-(u-1)}{(u-1)(u+1)}du$$ $$I=2u+\Big(\int\frac{du}{u-1}-\int\frac{du}{u+1}\Big)$$ $$I=2u+\ln|u-1|-\ln|u+1|+C$$ $$I=2u+\ln|\frac{u-1}{u+1}|+C$$ $$I=2\sqrt{x+1}+\ln|\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}|+C$$
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