Determine whether the improper integral is convergent or divergent. If it converges, evaluate it. $\int_{2}^{\infty} \frac{1}{x(\ln x)^{2}} d x$

Step 1 ：The given integral is an improper integral of type I, as the upper limit of integration is infinity. To determine whether it converges or diverges, we can directly compute the integral and see if it results in a finite number or not.

Step 2 ：We can solve the integral using the substitution method. Let \(u = \ln x\), then \(du = \frac{1}{x} dx\). The limits of integration will change accordingly: when \(x = 2\), \(u = \ln 2\); when \(x = \infty\), \(u = \infty\). The integral then becomes \(\int_{\ln 2}^{\infty} \frac{1}{u^{2}} du\), which is a standard integral and can be solved easily.

Step 3 ：The integral evaluates to a finite number, which means the original improper integral is convergent. The value of the integral is \(\frac{1}{\ln 2}\).

Step 4 ：Final Answer: The improper integral \(\int_{2}^{\infty} \frac{1}{x(\ln x)^{2}} d x\) is convergent and its value is \(\boxed{\frac{1}{\ln 2}}\).