# Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 461: 85

converges

#### Work Step by Step

Given $$\int_{8}^{\infty} \frac{d x}{x^{2}-4}$$ Since $x \geq 8\to \frac{1}{2} x^{2} \geq 4$, then \begin{aligned} -\frac{1}{2} x^{2}& \leq-4\\ \frac{1}{2} x^{2} &\leq x^{2}-4\\ \frac{1}{x^{2}-4} &\leq \frac{2}{x^{2}} \end{aligned} Since \begin{align*} \int_{8}^{\infty }\frac{2}{x^2}dx &= \lim_{R\to \infty } \int_{8}^{R }\frac{2}{x^2}dx\\ &=\frac{-2}{x}\bigg|_{8}^{\infty}\\ &= \frac{1}{4} \end{align*} Then by the comparison test, $\int_{8}^{\infty} \frac{d x}{x^{2}-4}$ also converges

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