Answer
$$y = \frac{3}{4}\ln \left| {\frac{{x - 4}}{x}} \right| + \ln \left| {4x - {x^2}} \right| + C$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{1 - 2x}}{{4x - {x^2}}} \cr
& dy = \frac{{1 - 2x}}{{4x - {x^2}}}dx \cr
& {\text{Completing the square}} \cr
& dy = \frac{{1 - 2x}}{{4x - {x^2} + 4 - 4}}dx \cr
& dy = \frac{{1 - 2x}}{{{{\left( {x - 2} \right)}^2} - 4}}dx \cr
& {\text{Let }}u = x - 2,{\text{ }}x = u + 2,{\text{ }}dx = du \cr
& dy = \frac{{1 - 2\left( {u + 2} \right)}}{{{u^2} - 4}}du \cr
& dy = \frac{{1 - 2u - 4}}{{{u^2} - 4}}du \cr
& y = \frac{3}{{{u^2} - 4}}du + \frac{{2u}}{{{u^2} - 4}}du \cr
& {\text{Integrating}} \cr
& y = \int {\frac{3}{{{u^2} - 4}}du} + \int {\frac{{2u}}{{{u^2} - 4}}} du \cr
& y = \frac{3}{{2\left( 2 \right)}}\ln \left| {\frac{{u - 2}}{{u + 2}}} \right| + \ln \left| {{u^2} - 4} \right| + C \cr
& {\text{Write in terms of }}x \cr
& y = \frac{3}{4}\ln \left| {\frac{{\left( {x - 2} \right) - 2}}{{\left( {x - 2} \right) + 2}}} \right| + \ln \left| {{{\left( {x - 2} \right)}^2} - 4} \right| + C \cr
& y = \frac{3}{4}\ln \left| {\frac{{x - 4}}{x}} \right| + \ln \left| {4x - {x^2}} \right| + C \cr} $$
