Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 65


$\frac{1}{\sqrt 3}\sec^{-1}(2x)+c$

Work Step by Step

Since $u=2x$, then $du= 2dx$. Now, we have $$\int \frac{ dx}{x\sqrt{12x^2-3}}=\frac{1}{\sqrt 3}\int \frac{ du}{u\sqrt{u^2-1}}\\=\frac{1}{\sqrt 3}\sec^{-1}u+c=\frac{1}{\sqrt 3}\sec^{-1}(2x)+c .$$
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