Answer
$$\frac{\pi }{3}$$
Work Step by Step
$$\eqalign{
& \int_2^4 {\frac{2}{{x\sqrt {{x^2} - 4} }}} d\theta \cr
& \frac{2}{{x\sqrt {{x^2} - 4} }}{\text{ has an infinite discontinuity at }}x = 2,{\text{so we can write}} \cr
& \int_2^4 {\frac{2}{{x\sqrt {{x^2} - 4} }}} d\theta = \mathop {\lim }\limits_{b \to {2^ + }} \int_b^4 {\frac{2}{{x\sqrt {{x^2} - 4} }}} dx \cr
& {\text{Integrate, use }}\int {\frac{1}{{x\sqrt {{x^2} - {a^2}} }}} dx = \frac{1}{a}\operatorname{arcsec} \left( {\frac{x}{a}} \right) + C \cr
& \mathop {\lim }\limits_{b \to {2^ + }} \int_b^4 {\frac{2}{{x\sqrt {{x^2} - 4} }}} d\theta = \mathop {\lim }\limits_{b \to {2^ + }} 2\left[ {\frac{1}{2}\operatorname{arcsec} \left( {\frac{x}{2}} \right)} \right]_b^4 \cr
& = \mathop {\lim }\limits_{b \to {2^ + }} \left[ {\operatorname{arcsec} \left( {\frac{x}{2}} \right)} \right]_b^4 \cr
& = \mathop {\lim }\limits_{b \to {2^ + }} \left[ {\operatorname{arcsec} \left( {\frac{4}{2}} \right) - \operatorname{arcsec} \left( {\frac{b}{2}} \right)} \right] \cr
& = \mathop {\lim }\limits_{b \to {2^ + }} \left[ {\frac{\pi }{3} - \operatorname{arcsec} \left( {\frac{b}{2}} \right)} \right] \cr
& {\text{Evaluate the limit when }}b \to {2^ + } \cr
& = \frac{\pi }{3} - \operatorname{arcsec} \left( {\frac{2}{2}} \right) \cr
& = \frac{\pi }{3} - 0 \cr
& = \frac{\pi }{3} \cr
& {\text{The following graph confirms the result}} \cr} $$
