Note that the function is defined for:
#x in (-oo,-sqrt6) uu (sqrt6,+oo)#
Consider the interval #x in (sqrt6,+oo)# and substitute:
#x =sqrt6cosht#
#dx =sqrt6 sinhtdt#
with #t > 0#
#int dx/(sqrt(x^2-6)) = sqrt6 int (sinhtdt)/(sqrt(6cosh^2t-6)) = int (sinhtdt)/(sqrt(cosh^2t-1) #
Use the hyperbolic identity:
#cosh^2t -1 = sinh^2t#
#int dx/(sqrt(x^2-6)) = int (sinhtdt)/(sqrt(sinh^2t) #
For #t > 0# we have that #sinht > 0#, so:
#int dx/(sqrt(x^2-6)) = int (sinhtdt)/sinht = int dt = t+C #
and undoing the substitution:
#int dx/(sqrt(x^2-6)) = "arcosh"(x/sqrt6) + C = ln (x/sqrt6 +sqrt (x^2/6-1)) +C#
#int dx/(sqrt(x^2-6)) = ln (x +sqrt (x^2-6)) +C#
Similarly in the interval #x in (-oo,-6)# we can substitute:
#x =-sqrt6cosht#
#dx =-sqrt6 sinhtdt#
with #t<0# and conclude that:
#int dx/(sqrt(x^2-6)) = ln abs(x +sqrt (x^2-6)) +C#
