Answer
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
