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: 90



Work Step by Step

Let $u=5x$, then $du=5dx$ and hence we have $$\int \frac{dx}{x\sqrt{25x^2-1}}=\int \frac{du}{u\sqrt{u^2-1}}\\ =\sec^{-1}u+c=\sec^{-1}(5x)+c.$$
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