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 442: 78


The integral $\int_{0}^{\infty} \dfrac{dx}{x^{3/2}(x+1)}$ diverges.

Work Step by Step

We are given the function $f(x)=\int_{0}^{\infty} \dfrac{dx}{x^{3/2}(x+1)}$ Since, $x^{5/2}\leq x^{3/2}$ This yields: $\dfrac{1}{x^{3/2}(x+1)} \geq \dfrac{1}{2x^{3/2}} $ Consider the integral $\int_{0}^{1} \dfrac{dx}{2x^{3/2}}=\dfrac{1}{2} [\dfrac{-2}{x^{1/2}}]_0^1 \\=\infty$ Thus, the integral $\int_{0}^{1} \dfrac{dx}{2x^{3/2}}$ diverges. Therefore, by the comparison test, the integral $\int_{0}^{\infty} \dfrac{dx}{x^{3/2}(x+1)}$ diverges as well.
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