Answer
$${\sin ^{ - 1}}\left( {\frac{{x - 2}}{2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{\sqrt {4x - {x^2}} }}} dx \cr
& {\text{Completing the square}} \cr
& 4x - {x^2} = - \left( {{x^2} - 4x} \right) \cr
& 4x - {x^2} = - \left( {{x^2} - 4x + 4 - 4} \right) \cr
& 4x - {x^2} = - \left[ {{{\left( {x - 2} \right)}^2} - 4} \right] = 4 - {\left( {x - 2} \right)^2},{\text{ then}} \cr
& \int {\frac{1}{{\sqrt {4x - {x^2}} }}} dx = \int {\frac{1}{{\sqrt {4 - {{\left( {x - 2} \right)}^2}} }}} dx \cr
& {\text{Let }}u = x - 2,{\text{ }}du = dx \cr
& \int {\frac{1}{{\sqrt {4 - {{\left( {x - 2} \right)}^2}} }}} dx = \int {\frac{{du}}{{\sqrt {4 - {u^2}} }}} \cr
& {\text{Apply the formula }}\int {\frac{1}{{\sqrt {{a^2} - {x^2}} }}dx = {{\sin }^{ - 1}}\left( {\frac{x}{a}} \right) + C,{\text{ then}}} \cr
& \int {\frac{{du}}{{\sqrt {4 - {u^2}} }}} = {\sin ^{ - 1}}\left( {\frac{u}{2}} \right) + C \cr
& {\text{Substitute back }}u = x - 2 \cr
& = {\sin ^{ - 1}}\left( {\frac{{x - 2}}{2}} \right) + C \cr} $$
