`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`

`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`