We begin by completing the square of the denominator.
#x^2-16x+64+37-64#
#x^2-16x+64-27#
#(x-8)^2-27#
The integral becomes
#int 1/sqrt((x-8)^2-27)dx#
Now we choose a substitution
Let #sec\theta=(x-8)/sqrt(27)#
Solve for #x#
#x-8=sqrt(27)sec\theta#
#x=sqrt(27)sec\theta+8#
Differentiate both sides
#dx=sqrt(27)sec\thetatan\thetad\theta#
Now make the substitution into the integral
#int (sqrt(27)sec\thetatan\theta)/(sqrt((sqrt(27)sec\theta+8-8)^2-27))d\theta#
#int (sqrt(27)sec\thetatan\theta)/(sqrt(27sec^2\theta-27))d\theta#
#int (sqrt(27)sec\thetatan\theta)/(sqrt(27(sec^2\theta-1)))d\theta#
#sqrt(27)/sqrt(27)int (sec\thetatan\theta)/(sqrt(tan^2\theta))d\theta#
#int (sec\thetatan\theta)/(tan\theta)d\theta#
#int sec\thetad\theta#
Now we can integrate
#ln|sec\theta+tan\theta|#
We have to get things back in terms of #x#
#sec\theta=(x-8)/sqrt(27)#
#tan\theta=(sqrt((x-8)^2-27))/sqrt(27)#
Back substituting we have
#ln|(x-8+sqrt((x-8)^2-27))/sqrt(27)|+C#
Using properties of logarithms we can rewrite
#ln|x-8+sqrt((x-8)^2-27)|-ln|sqrt(27)|+C#
#ln|sqrt(27)|# is just a constant so we can write
#ln|x-8+sqrt((x-8)^2-27)|+C#
You may also see the answer written as follows
#ln|x-8+sqrt(x^2-16x+37)|+C#
