Use the substitution #x = 3tantheta#. Then #dx = 3sec^2theta d theta#.
#=>int (3tantheta)^2/sqrt((3tantheta)^2 + 9) * 3sec^2theta d theta#
#=>int (9tan^2theta)/sqrt(9tan^2theta +9) * 3sec^2theta d theta#
#=>int (9tan^2theta)/sqrt(9sec^2theta) * 3sec^2theta d theta#
#=>int (9tan^2theta)/(3sectheta) * 3sec^2theta d theta#
#=>int 9tan^2thetasectheta d theta#
#=>9int tantheta(tanthetasectheta)d theta#
Use integration by parts for this one. Let #u = tantheta# and #dv = tanthetasectheta d theta#. Then #du = sec^2thetad theta# and #v = sectheta#.
#=>9tanthetasectheta - 9intsec^3theta#
This is a (relatively) known integral. The proof can be found here
#=>9tanthetasectheta - 9/2secthetatantheta - 9/2ln|sectheta + tantheta| + C#
#=>9/2tanthetasectheta - 9/2ln|sectheta + tantheta| + C#
We know from our initial substitution that #tantheta = x/3#. This means that the hypotenuse of the right triangle would be #sqrt(x^2 + 9)#, and that #sectheta = sqrt(x^2 + 9)/3#.
#=>9/2(x/3)sqrt(x^2 + 9)/3 - 9/2ln|sqrt(x^2 + 9)/3 + x/3| + C#
#=>(xsqrt(x^2 + 9))/2 - 9/2ln|(x + sqrt(x^2 +9))/3| + C#
Hopefully this helps!
