Thomas' Calculus 13th Edition

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

Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 286: 32



Work Step by Step

$\int^{\frac{1}{\sqrt{3}}}_{0}\frac{1}{1-4x^2}dx$ =$\int^{\frac{1}{\sqrt{3}}}_{0}\frac{1}{1-2^2x^2}dx$ =$\frac{1}{2}tan^{-1}(\frac{2}{\sqrt{3}})-\frac{1}{2}tan^{-1}(0)$ =$\frac{1}{2}tan^{-1}(\frac{2}{\sqrt{3}})$
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.