By using the substitution #x=3sec theta#,
#int{sqrt{x^2-9}}/{x^3}dx=1/6sec^{-1}(x/3)-{sqrt{x^2-9}}/{2x^2}+C#.
Let #x=3sec theta Rightarrow dx=3sec theta tan theta d theta#.
#int{sqrt{x^2-9}}/{x^3}dx=int{sqrt{(3sec theta)^2 -9}}/{(3sec theta)^3}3sec theta tan theta d theta#,
which simplifies to
#=1/3int{tan^2 theta}/{sec^2 theta}d theta#
by using #tan theta={sin theta}/{cos theta}# and #sec theta=1/{cos theta}#,
#=1/3int sin^2 theta d theta#
by using #sin^2 theta ={1-cos(2theta)}/2#,
#1/6int[1-cos(2theta)]d theta=1/6[theta-sin(2theta)/2]+C#
by using #sin(2theta)=2sin theta cos theta#,
#=1/6(theta-sin theta cos theta)+C#
Now, we need to rewrite our answer above in terms of #x#.
#x=3sec theta Rightarrow x/3=sec theta Rightarrow theta=sec^{-1}(x/3)#
Since #sec theta=x/3#, we can construct a right triangle with angle #theta# such that its hypotenuse is #x#, its adjacent is #3#, and its opposite is #sqrt{x^2-9}#. So, we have
#sin theta={sqrt{x^2-9}}/{x}# and #cos theta=3/x#
Hence, by rewriting #theta#, #sin theta#, #cos theta# in terms of #x#,
#int{sqrt{x^2-9}}/{x^3}dx=1/6sec^{-1}(x/3)-{sqrt{x^2-9}}/{2x^2}+C#.
