The integral $$\int_{2}^{\,\infty} {\frac{3}{x\sqrt{x^2-4}}}\, dx$$ is improper for two reasons: the interval $[2,\infty]$ is infinite and the integrand has an infinite discontinuity at $x = 2$. Evaluate it by expressing it as a sum of improper integrals of Type 2 and Type 1 as follows: $$\int_{2}^{\,\infty} {\frac{3}{x\sqrt{x^2-4}}}\, dx = \int_{2}^{\,b} {\frac{3}{x\sqrt{x^2-4}}}\, dx + \int_{b}^{\,\infty} {\frac{3}{x\sqrt{x^2-4}}}\, dx,$$ where $b$ is any finite number greater than 2.

If the improper integral diverges, type an upper-case "D".