Complete the square at the denominator:
#int dx/sqrt(9x^2-6x-3) = int dx/sqrt( (3x-1)^2 -4) = 1/2 int dx/sqrt( ((3x-1)/2)^2 -1)#
Note that the integrand is defined only for:
#abs (3x-1) > 2#
that is for #x > 1# or #x <-1/3#.
Let #x > 1# and substitute:
#(3x-1)/2 = sect#
#dx =2/3 sect tant dt#
with #t in (0,pi/2)#. Then
#int dx/sqrt(9x^2-6x-3) = 1/3 int (sect tantdt)/sqrt(sec^2t-1)#
Using the trigonometric identity:
#sec^2t -1 = tan^2t#
and considering that for #t in (0,pi/2)# the tangent is positive:
#int dx/sqrt(9x^2-6x-3) = 1/3 int (sect tantdt)/sqrt(tan^2t)#
#int dx/sqrt(9x^2-6x-3) = 1/3 int (sect tantdt)/tant#
#int dx/sqrt(9x^2-6x-3) = 1/3 int sect dt#
#int dx/sqrt(9x^2-6x-3) = 1/3 ln abs ( sect +tant)+C#
#int dx/sqrt(9x^2-6x-3) = 1/3 ln abs ( (3x-1)/2 +sqrt(((3x-1)/2)^2-1))+C#
Multiplying the argument of the logarithm by #2# which is equivalent to adding a constant:
#int dx/sqrt(9x^2-6x-3) = 1/3 ln abs ( 3x-1 +sqrt(9x^2-6x-3))+C#
and by direct differentiation we can see that this is valid also for #x < -1/3#
