#int1/sqrt(x^2-4)dx = int1/sqrt(4(x^2/4-1))dx#
#=1/2int1/sqrt((x/2)^2-1)dx#
Let #x/2 = sec(theta)#
Then #dx = 2sec(theta)tan(theta)#
#=>int1/sqrt(x^2-4)dx = 1/2int1/sqrt(sec^2(theta)-1)*2sec(theta)tan(theta)d theta#
#=int (sec(theta)tan(theta))/sqrt(tan^2(theta))d theta#
#=int sec(theta)d theta#
#=ln|sec(theta) + tan(theta)| + C#
#=ln|x/2 + sqrt(x^2/4-1)|+C#
(For the final equality, try drawing a right triangle with angle #theta# such that #sec(theta) = x/2# and solve for #tan(theta)#)
We could stop here, but we can also make this look a little nicer by getting rid of the #1/2#
#ln|x/2+sqrt(x^2/4-1)|+C =ln|1/2(x+2sqrt(x^2/4-1))|+C#
#=ln|x+sqrt(x^2-4)|+ln(1/2)+C#
But as #C# is an arbitrary constant, we can include the #ln(1/2)# in that to give us the final result
#ln|x+sqrt(x^2-4)|+C#