I suppose you mean to have the entire numerator under the square root?
[tex]\displaystyle\int_2^4\frac{\sqrt{x^2-4}}{x^2}\,\mathrm dx[/tex]
We can use a trigonometric substitution to start:
[tex]x=2\sec t\implies\mathrm dx=2\sec t\tan t\,\mathrm dt[/tex]
Then for [tex]x=2[/tex], [tex]t=\sec^{-1}1=0[/tex]; for [tex]x=4[/tex], [tex]t=\sec^{-1}2=\frac\pi3[/tex]. So the integral is equivalent to
[tex]\displaystyle\int_0^{\pi/3}\frac{\sqrt{(2\sec t)^2-4}}{(2\sec t)^2}(2\sec t\tan t)\,\mathrm dt=\int_0^{\pi/3}\frac{\tan^2t}{\sec t}\,\mathrm dt[/tex]
We can write
[tex]\dfrac{\tan^2t}{\sec t}=\dfrac{\frac{\sin^2t}{\cos^2t}}{\frac1{\cos t}}=\dfrac{\sin^2t}{\cos t}=\dfrac{1-\cos^2t}{\cos t}=\sec t-\cos t[/tex]
so the integral becomes
[tex]\displaystyle\int_0^{\pi/3}(\sec t-\cos t)\,\mathrm dt=\boxed{\ln(2+\sqrt3)-\frac{\sqrt3}2}[/tex]
