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