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: 20


$(\dfrac{1}{6}) \ln |t^2-2|-(\dfrac{1}{6}) \ln |t^2+1|+c$

Work Step by Step

Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$ Solve $\int \dfrac{1}{t^4-t^2-2}dt$ Write the integral into partial fractions as follows: $\int \dfrac{1}{t^4-t^2-2}dt=\int [(\dfrac{1}{3}) \dfrac{t}{t^2-2}-(\dfrac{1}{3}) \dfrac{t}{t^2+1}] dt$ Let us consider $u=(t^2-2) \implies du=(2t) dt$ and $a=t^2+1 \implies da=(2t) dt$ This implies that $\int (\dfrac{1}{6}) \dfrac{du}{u}-\int(\dfrac{1}{6}) \dfrac{da}{a}=(\dfrac{1}{6})\ln |u|-(\dfrac{1}{6})\ln |a|=(\dfrac{1}{6}) \ln |t^2-2|-(\dfrac{1}{6}) \ln |t^2+1|+c$
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