Let
#I = intdx/sqrt(9x-12sqrtx-1) #
We want to get rid of the square roots in the square roots, so let #u = sqrtx#. Therefore,
#I = int (2udu)/sqrt(9u^2-12u-1) = int\ (2udu)/sqrt((3u-2)^2-5) #
To simplify, let #v = 3u-2#. Therefore,
#I = 2/9 int\ \ ((v+2)dv)/sqrt(v^2-5) = 2/9 int (vdv)/sqrt(v^2-5) + 4/9 int (dv)/sqrt(v^2-5)#
Now we have some tools for things like this, but that 5 is pesky, so let's let #y = v / sqrt(5)#, such that
#I = 2/9 int (5ydy)/sqrt(5(y^2-1)) + 4/9 int (dy sqrt(5))/sqrt(5(y^2-1)) #
#= (2sqrt(5))/9 int (ydy)/sqrt(y^2-1) + 4/9 int dy/sqrt(y^2-1) #
These two sides now will have different methods. The left side is a simple u-subsitution, with #w = y^2#, and the right with a more complicated trig substitution with #y = sec z#. Therefore,
#= sqrt(5)/9 int (dw)/sqrt(w-1) + 4/9 int\ \ secz\ dz#
Finally, using #g = w-1# and #\Omega = secz + tanz# (this final substitution is famous, I can't really give any logical jump to easily deduce it):
#(2sqrt5)/9 * int\ 1/2 g^(-1/2)dg + 4/9 int\ (dOmega)/Omega #
So
#I = (2sqrt5)/9 sqrtg + 4/9 lnOmega + C#
for some constant #C#, and where
#sqrt(g) = sqrt(w-1) = sqrt(y^2-1) = sqrt(v^2/5 - 1) = 5^(-1/2) sqrt(v^2-5) #
#= 1/sqrt5 sqrt((3u-2)^2-5) = 1/sqrt(5) * sqrt(9x - 12 sqrtx - 1) #
and
#Omega = secz + tan z = secz + sqrt(sec^2z - 1) = y + sqrt(y^2-1) #
#= 1/sqrt5 (v+sqrt(v^2-5)) = 1/sqrt5(3sqrtx - 2 + sqrt(9x - 12sqrtx - 1)) #
