Answer
$$
\int_{2}^{\infty} \frac{d x}{x\sqrt{x^{2}-4}}
$$
The given improper integral is convergent and its value is
$$
\int_{2}^{\infty} \frac{d x}{x\sqrt{x^{2}-4}}=\frac{\pi}{4}
$$
Work Step by Step
$$
\int_{2}^{\infty} \frac{d x}{x\sqrt{x^{2}-4}}
$$
The given improper integral can be expressed as a sum of improper integrals of Type 2 and Type 1 as follows:
$$
\int_{2}^{\infty} \frac{d x}{x\sqrt{x^{2}-4}}=\int_{2}^{3} \frac{d x}{x\sqrt{x^{2}-4}}+\int_{3}^{\infty} \frac{d x}{x\sqrt{x^{2}-4}}
$$
Observe that the first integral is improper because $ f(x)= \frac{1}{x\sqrt{x^{2}-4}} $ has the vertical asymptote $ x =2 $. Since the infinite discontinuity occurs at the left endpoint of $ [2,3] $
$$
\begin{split}
\int_{2}^{\infty} \frac{d x}{x\sqrt{x^{2}-4}} &=\int_{2}^{3} \frac{d x}{x\sqrt{x^{2}-4}}+\int_{3}^{\infty} \frac{d x}{x\sqrt{x^{2}-4}}\\
&=\lim _{t \rightarrow 2^{+}} \int_{t}^{3} \frac{d x}{x\sqrt{x^{2}-4}} +\lim _{t \rightarrow \infty}\int_{3}^{t} \frac{d x}{x\sqrt{x^{2}-4}} \quad (1)
\end{split}
$$
Now
$$
\begin{split}
\int \frac{d x}{x\sqrt{x^{2}-4}}&=\int \frac{d x}{x\sqrt{x^{2}-4}} \\
& \quad\quad\left[\begin{array}{c}{x=2\sec u, \text {where }} \\ {0 \leq u \lt \frac{\pi}{2}, \text {or } \pi \leq u \lt \frac{3\pi}{2}}\end{array}\right] \\
& =\int \frac{2 \sec u \tan u d u}{2\sec u. 2 \tan u} \\
& =\frac{1}{2}u +C\\
&=\frac{1}{2} \sec^{-1}(\frac{1}{2}x) +C \\
\end{split}
$$
substituting in Eq. (1 ) we obtain that, $$
\begin{split}
\int_{2}^{\infty} \frac{d x}{x \sqrt{x^{2}-4}} &=\lim _{t \rightarrow 2^{+}}\left[\frac{1}{2} \sec ^{-1}\left(\frac{1}{2} x\right)\right]_{t}^{3}+\lim _{t \rightarrow \infty}\left[\frac{1}{2} \sec ^{-1}\left(\frac{1}{2} x\right)\right]_{3}^{t} \\
&=\frac{1}{2} \sec ^{-1}\left(\frac{3}{2}\right)-0+\frac{1}{2}\left(\frac{\pi}{2}\right)-\frac{1}{2} \sec ^{-1}\left(\frac{3}{2}\right) \\
&=\frac{\pi}{4}.
\end{split}
$$
The given improper integral is convergent and its value is
$$
\int_{2}^{\infty} \frac{d x}{x\sqrt{x^{2}-4}}=\frac{\pi}{4}.
$$
