Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.8 - Improper Integrals - Exercises 8.8 - Page 501: 12


$\ln 3$

Work Step by Step

Consider $f(t)= \int_2^{\infty} \dfrac{2 dt}{t^2-1}$ This can be re-written as: $\int_2^{\infty} \dfrac{2 dt}{(t^2-1)}=\int_2^{\infty} \dfrac{2 dt}{(t-1)(t+1)}$ Now, $\int_2^{\infty} \dfrac{1}{(t-1)} dt-\int_2^{\infty} \dfrac{1}{(t+1)} dt=\lim\limits_{k \to \infty} [\ln |\dfrac{t-1}{t+1}|]_2^{k} $ and $ \lim\limits_{k\to \infty} [\ln |\dfrac{1-\dfrac{1}{k}}{1+\dfrac{1}{k}}|]-\ln (\dfrac{1}{3}) $ or, $\ln |\dfrac{1-0}{1+0}|-\ln (\dfrac{1}{3}) =-\ln (\dfrac{1}{3}) =\ln 3$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.