Use trig substitution. Let #x= 9tantheta#. Then #dx = 9sec^2theta d theta#.
#=>intsqrt((9tantheta)^2 + 81) * 9sec^2theta d theta#
#=>intsqrt(81tan^2theta + 81) * 9sec^2theta d theta#
#=>intsqrt(81(tan^2theta +1)) * 9sec^2theta d theta#
#=>intsqrt(81sec^2theta) * 9sec^2theta d theta#
#=>int 9sectheta * 9sec^2theta d theta#
#=>int 81sec^3theta d theta#
#=>81int sec^3theta d theta#
This is a relatively known integral. The proof can be found here.
#=>81/2secthetatantheta+ 81/2ln|sectheta+ tantheta| + C#
We know from our initial substitution that #tantheta = x/9#. This means the hypotenuse of the triangle would have a hypotenuse of #sqrt(x^2 + 81)#.
#=>81/2(sqrt(x^2 + 81)/9)(x/9) + 81/2ln|sqrt(x^2 + 81)/9 + x/9| + C#
#=>(xsqrt(x^2 + 81))/2 + 81/2ln|(x + sqrt(x^2 + 81))/9| + C#
Hopefully this helps!
