Use the identity #sec^2u - 1 -= tan^2u#.
Substitute #x = 7secu#.
For #x>7#, let #0<\u<\pi/2#.
#frac{dx}{du} = 7 secu tanu#
#int 1/sqrt{x^2 - 49} dx = int 1/sqrt{x^2 - 49} frac{dx}{du} du #
#= int 1/sqrt{(7secu)^2 - 49} (7 secu tanu) du#
#= int frac{7 secu tanu}{7sqrt{sec^2u - 1}} du#
#= int secu frac{tanu}{sqrt{tan^2u}} du#
Since #0<\u<\pi/2#, #sqrt{tan^2u} -= tanu#.
#int secu frac{tanu}{sqrt{tan^2u}} du = int secu du#
#= ln|secu+tanu| + C_1#, where #C_1# is an integration constant.
#= ln(secu+tanu) + C_1#
#= ln(7secu+7tanu) + C_2#, where #C_2=C_1-ln7#.
#= ln(x + sqrt{x^2 - 49}) + C_2#
For #x<-7#, let #pi/2<\u<\pi#, then #sqrt{tan^2u} -= -tanu#.
#int secu frac{tanu}{sqrt{tan^2u}} du = -int secu du#
#= -ln|secu+tanu| + C_3#, where #C_3# is an integration constant.
#= -ln(-secu-tanu) + C_3#
#= -ln(-7secu-7tanu) + C_4#, where #C_4=C_3+ln7#.
#= -ln(-x+sqrt{x^2-49}) + C_4#