To integrate this function we will use a trigonometric substitution.
Let #costheta=3/x#
Therefore, #sectheta=x/3# and #x =3sectheta#
Differentiate #x=3sectheta#
#dx=3secthetatanthetad theta#
Make the substitution into the integral
#intsqrt((3sectheta)^2-9)/(3sectheta)3secthetatanthetad theta#
#intsqrt(9sec^2theta-9)tantheta d theta#
#intsqrt(9(sec^2theta-1))tanthetad theta#
#intsqrt(9)sqrt(tan^2theta)tanthetad theta#
#3inttan^2thetad theta#
#3intsec^2theta-1d theta#
Now integrating we get
#3(tantheta-theta)#
Now define #theta# and #tantheta # in terms of #x# as follows
#theta=arctan(sqrt(x^2-9)/3) #
#tantheta = sqrt(x^2-9)/3 #
now back substitute
#3(sqrt(x^2-9)/3-arctan(sqrt(x^2-9)/3))+C #
Distributing the #3# we will have
#sqrt(x^2-9)-3arctan(sqrt(x^2-9)/3)+C # FINAL ANSWER
