Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.7 Improper Integrals - Exercises - Page 441: 63


The integral $\int_{3}^{\infty} \dfrac{d x}{\sqrt x-1}$ diverges.

Work Step by Step

We are given the function $f(x)=\int_{3}^{\infty} \dfrac{d x}{\sqrt x-1} $ Since, $\sqrt x \geq \sqrt x-1$, for $x \gt 1$, we have: $\dfrac{1}{\sqrt x} \leq \dfrac{1}{\sqrt x-1}$ The integral $\int_{1}^{\infty} \dfrac{d x}{\sqrt x}=\int_{1}^{\infty} \dfrac{d x}{x^{1/2}} $ diverges because $\dfrac{1}{2} \lt 1$ Therefore, by the comparison test, the integral $\int_{3}^{\infty} \dfrac{d x}{\sqrt x-1}$ diverges.
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