Answer
$$\frac{1}{4}\ln \left| {\frac{{x - 4}}{x}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{ - 1}}{{4x - {x^2}}}} dx \cr
& {\text{Completing the square}} \cr
& 4x - {x^2} = - \left( {{x^2} - 4x + 4 - 4} \right) \cr
& 4x - {x^2} = - \left[ {\left( {{x^2} - 4x + 4} \right) - 4} \right] = - \left[ {{{\left( {x - 2} \right)}^2} - 4} \right] \cr
& 4x - {x^2} = 4 - {\left( {x - 2} \right)^2} \cr
& {\text{Then,}} \cr
& \int {\frac{{ - 1}}{{4x - {x^2}}}} dx = \int {\frac{{ - 1}}{{4 - {{\left( {x - 2} \right)}^2}}}} dx \cr
& = \int {\frac{1}{{{{\left( {x - 2} \right)}^2} - 4}}} dx \cr
& {\text{Let }}u = x - 2,{\text{ }}du = dx \cr
& = \int {\frac{1}{{{u^2} - 4}}} du \cr
& {\text{Use Theorem 5}}{\text{.20 }}\int {\frac{{du}}{{{u^2} - {a^2}}} = } \frac{1}{{2a}}\ln \left| {\frac{{u - a}}{{u + a}}} \right| + C \cr
& \int {\frac{1}{{{u^2} - 4}}} du = \frac{1}{{2\left( 2 \right)}}\ln \left| {\frac{{u - 2}}{{u + 2}}} \right| + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{4}\ln \left| {\frac{{\left( {x - 2} \right) - 2}}{{\left( {x - 2} \right) + 2}}} \right| + C \cr
& = \frac{1}{4}\ln \left| {\frac{{x - 4}}{x}} \right| + C \cr} $$
