The integrand is defined for #x in (-oo,-a) uu (a,+oo)#. Let us focus first on #x in (a,+oo)# and substitute:
#x = a sect#
#dx = asect tant#
with #t in (0,pi/2)#
so:
#int dx/sqrt(x^2-a^2) = int (asect tant dt)/sqrt(a^2sec^2t-a^2)#
#int dx/sqrt(x^2-a^2) = int (sect tant dt)/sqrt(sec^2t-1)#
Use now the trigonometric identity:
#sec^2t-1 = tan^2t#
and as for #t in (0,pi/2)# the tangent is positive:
#sqrt(sec^2t-1) = tan^#
Then:
#int dx/sqrt(x^2-a^2) = int (sect tant dt)/tant#
#int dx/sqrt(x^2-a^2) = int sect dt#
#int dx/sqrt(x^2-a^2) = ln abs (sect +tant) +C#
Undoing the substitution:
#int dx/sqrt(x^2-a^2) = ln abs (x/a +sqrt(x^2/a^2-1)) +C#
#int dx/sqrt(x^2-a^2) = ln abs (x +sqrt(x^2-a^2)) +C#
By differentiating we can see that the solution is valid also for #x in (-oo,-3)#