Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 13 - Section 13.2 - Derivatives and Integrals of Vector Functions - 13.2 Exercise - Page 861: 37


$\int_0^1\langle\frac{1}{t+1},\frac{1}{t^2+1},\frac{t}{t^2+1}\rangle dt=\langle\ln{2},\frac{\pi}{4},\frac{\ln2}{2}\rangle$.

Work Step by Step

$\int_0^1\langle\frac{1}{t+1},\frac{1}{t^2+1},\frac{t}{t^2+1}\rangle dt=\int_0^1\frac{1}{t+1}dt+\int_0^1\frac{1}{t^2+1}dt+\int_0^1\frac{t}{t^2+1}dt=(\ln{(t+1)})\big|_0^1+(\arctan{t})\big|_0^1+(\frac{\ln{{(t^2+1)}}}{2})\big|_0^1=\langle\ln{2},\frac{\pi}{4},\frac{\ln2}{2}\rangle$
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