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