Answer
$$y = - \frac{1}{{\sqrt 3 }}\ln \left| {\frac{{\sqrt 3 + \sqrt { - 4{x^2} + 8x - 1} }}{{2\left( {x - 1} \right)}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{1}{{\left( {x - 1} \right)\sqrt { - 4{x^2} + 8x - 1} }} \cr
& {\text{Completing the square}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\left( {x - 1} \right)\sqrt { - \left( {4{x^2} - 8x + 1} \right)} }} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\left( {x - 1} \right)\sqrt {3 - \left( {4{x^2} - 8x + 4} \right)} }} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\left( {x - 1} \right)\sqrt {3 - {{\left( {2x - 2} \right)}^2}} }} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\left( {x - 1} \right)\sqrt {{{\left( {\sqrt 3 } \right)}^2} - {{\left[ {2\left( {x - 1} \right)} \right]}^2}} }} \cr
& {\text{Let }}u = 2\left( {x - 1} \right),{\text{ }}du = 2dx,{\text{ }}dx = \frac{1}{2}du \cr
& dy = \frac{1}{{\left( {u/2} \right)\sqrt {{{\left( {\sqrt 3 } \right)}^2} - {u^2}} }}\left( {\frac{1}{2}} \right)du \cr
& {\text{Integrate both sides}} \cr
& \int {dy} = \int {\frac{1}{{u\sqrt {{{\left( {\sqrt 3 } \right)}^2} - {u^2}} }}} du \cr
& {\text{By theorem 5}}{\text{.20}} \cr
& y = - \frac{1}{{\sqrt 3 }}\ln \left| {\frac{{\sqrt 3 + \sqrt {3 - {u^2}} }}{u}} \right| + C \cr
& {\text{Substitute }}u = 2\left( {x - 1} \right) \cr
& y = - \frac{1}{{\sqrt 3 }}\ln \left| {\frac{{\sqrt 3 + \sqrt {3 - {{\left( {2x - 2} \right)}^2}} }}{u}} \right| + C \cr
& y = - \frac{1}{{\sqrt 3 }}\ln \left| {\frac{{\sqrt 3 + \sqrt { - 4{x^2} + 8x - 1} }}{{2\left( {x - 1} \right)}}} \right| + C \cr} $$
