int 1/((x-1)sqrt(4x^2-8x+3)) dx Find the indefinite integral

int1/((x-1)sqrt(4x^2-8x+3))dx
Let's rewrite the integral by completing the square of the term in denominator,
=int1/((x-1)sqrt((2x-2)^2-1))dx
Apply the integral substitution: u=2x-2
du=2dx
=>dx=(du)/2
u=2(x-1)
=>(x-1)=u/2
=int1/((u/2)sqrt(u^2-1))(du)/2
=int1/(usqrt(u^2-1))du
Again apply integral substitution: u=sec(v)
du=sec(v)tan(v)dv
=int1/(sec(v)sqrt(sec^2(v)-1))sec(v)tan(v)dv
=inttan(v)/(sqrt(sec^2(v)-1))dv
Use the trigonometric identity:sec^2(x)=1+tan^2(x)
=inttan(v)/sqrt(1+tan^2(v)-1)dv
=inttan(v)/sqrt(tan^2(v))dv
=inttan(v)/tan(v)dv   assuming tan(v) >=0
=intdv
=v
Substitute back v=arcsec(u)  and  u=(2x-2)
and add a constant C to the solution,
=arcsec(2x-2)+C