# The value of the integral is:

$\hspace{-0.6 cm }Let \bf{I = \int_{2}^{4}\frac{\sqrt{x^2-4}}{x^4}dx\;,} Now Put \bf{x=\frac{1}{t}} and \bf{dx = -\frac{1}{t^2}dt} and changing\\\\ \\limits, We get \bf{I = \int_{\frac{1}{4}}^{\frac{1}{2}}t\sqrt{1-4t^2}dt\;,} Now Put \bf{(1-4t^2) = v\;, } Then \bf{-8tdt = dv}\\\\\\ And changing limit, So we get \bf{I = \frac{1}{8}\int_{0}^{\frac{3}{4}}\sqrt{v}dv = = \frac{1}{12}\left[v\sqrt{v}\right]_{0}^{\frac{3}{4}} = \frac{3}{32}.}$