<s>[INST] Determine whether $\int^{\infty}_{2} \frac{1}{x\sqrt{x^2-4}} dx$ converges. Determine whether $$\int^{\infty}_{2} \frac{1}{x\sqrt{x^2-4}} dx$$ converges.
Doing some rough work, I realize that this function near $\infty$, behaves like $$\frac{1}{x\sqrt{x^2}} = \frac{1}{x^2}$$
I know this function converges, but I am having a hard time finding a $larger$ function that converges too.
I know that:
$$\frac{1}{x\sqrt{x^2-4}} \leq \frac{1}{\sqrt{x^2-4}}$$
but the bigger function diverges, so it doesn't work.
 [/INST] Let $x-2=u$. Then $dx=du$. The integral becomes \begin{align*} \int_0^\infty \frac{du}{(u+2)\sqrt{(u+2)^2-4}}&=\int_0^\infty \frac{du}{(u+2)\sqrt{u^2+4u}}=\int_0^\infty \frac{du}{\sqrt{u^3+6u^2+8u}} \\ \\ &= \underbrace{\int_0^1 \frac{du}{\sqrt{u^3+6u^2+8u}}}_{=I_1} +\underbrace{\int_1^\infty \frac{du}{\sqrt{u^3+6u^2+8u}}}_{=I_2} \end{align*}
$\bullet$ Now, since $\displaystyle \frac{1}{\sqrt{u^3+6u^2+8u}} \geq \frac{1}{\sqrt{u^3}} = \frac{1}{u^{3/2}}$, $\ I_2$ converges by $p-$test ($3/2>1)$.
$\bullet$ Since, $\displaystyle \frac{1}{\sqrt{u^3+6u^2+8u}} \geq \frac{1}{\sqrt{8u}} = \frac{1}{\sqrt 8 u^{1/2}}$, $ \ \displaystyle \frac{I_1}{\sqrt 8}$ converges and hence $I_1$ converges by 
$p-$test ($1/2<1)$.
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