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: 10


$\dfrac{-\ln |x+1|}{2}+\dfrac{3\ln |x+3|}{2} +c$

Work Step by Step

Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$ $\int \dfrac{x}{x^2+4x+3}=\int \dfrac{x}{(x+1)(x+3)}$ Write the integral into partial fractions as follows: $\int \dfrac{x}{(x+1)(x+3)}=\int [-\dfrac{1}{2(x+1)}+\dfrac{3}{2(x+3)}]dx$ Hence, $\int [-\dfrac{1}{2(x+1)}+\dfrac{3}{2(x+3)}]dx=\dfrac{-\ln |x+1|}{2}+\dfrac{3\ln |x+3|}{2} +c$
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