The substitution: x = 4 sec t ( or the Greek letter: theta )
dx = 4 tan t sec t
\(\int { \frac{1}{ x^{2} \sqrt{ x^{2} -16} } } \, dx= \\ = \int { \frac{4 tan t sect}{16 sect \sqrt{16 sec^{2}t-16 } } } \, dt= \\ \int { \frac{4tant}{16sect*4tant} } \, dt = \\ = \frac{1}{16} \int { cos t} \, dt = \frac{1}{16}sin t\)
From the substitution we have: cos x = 4 / x
sin² x = 1 - 16/x²
sin x =\(\frac{ \sqrt{ x^{2} -16} }{x}\)
Final solution is: \(\frac{1}{16} * \frac{ \sqrt{ x^{2} -16} }{x} +C\)
