University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Practice Exercises - Page 474: 11

Answer

$\ln |\dfrac{x}{x+1}|+\dfrac{1}{x+1} +c$

Work Step by Step

Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$ Write the integral into partial fractions as follows: $\int \dfrac{dx}{x(x+1)^2}=\int [\dfrac{1}{x}-\dfrac{1}{x+1}-\dfrac{1}{(x+1)^2} ]dx$ Hence, $\ln |x|-\ln |x+1|+\dfrac{1}{x+1}+c=\ln |\dfrac{x}{x+1}|+\dfrac{1}{x+1} +c$
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