Answer
$$\displaystyle{\int\frac{1}{t^2\sqrt{t^2-16}}dt=\frac{\sqrt{t^2-16}}{ 16t}+C}$$
Work Step by Step
$\displaystyle{I=\int\frac{1}{t^2\sqrt{t^2-16}}dt}\\$
$\displaystyle \left[\begin{array}{ll} t=4\sec\theta & t^2=16\sec^2\theta \\ & \\ \frac{dt}{d\theta}=4\sec\theta\tan\theta & dt=4\sec\theta\tan\theta\ d\theta \end{array}\right]$ Integration by substitution
$\displaystyle{I=\int\frac{1}{16\sec^2\theta\sqrt{16\sec^2\theta-16}}4\sec\theta\tan\theta\ d\theta}\\
\displaystyle{I=\int\frac{1}{4\sec\theta\sqrt{16\left(\sec^2\theta-1\right)}}\tan\theta\ d\theta}\\
\displaystyle{I=\int\frac{1}{16\sec\theta}\ d\theta}\\
\displaystyle{I=\frac{1}{16}\int\cos\theta\ d\theta}\\
\displaystyle{I=\frac{1}{16}\sin\theta+C}\\$
$\sin\theta=\frac{\sqrt{t^2-16}}{ t}\\$
$\displaystyle{I=\frac{\sqrt{t^2-16}}{ 16t}+C}\\$