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: 19


$(\dfrac{1}{2}) \tan^{-1} t-\dfrac{\sqrt 3}{6}\tan^{-1} (\dfrac{t}{\sqrt 3})+c$

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 $\int \dfrac{1}{t^4+4t^2+3}dt$ Write the integral into partial fractions as follows: $\int \dfrac{1}{t^4+4t^2+3}dt=\int [(\dfrac{1}{2}) \dfrac{1}{t^2+1}-(\dfrac{1}{2}) \dfrac{1}{t^2+3}] dt$ This implies that $(\dfrac{1}{2}) \tan^{-1} t-\dfrac{1}{2\sqrt 3}\tan^{-1} (\dfrac{t}{\sqrt 3})+c=(\dfrac{1}{2}) \tan^{-1} t-\dfrac{\sqrt 3}{6}\tan^{-1} (\dfrac{t}{\sqrt 3})+c$
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