Answer
$$ - 2\sqrt 3 + \frac{1}{6}\pi + 4$$
Work Step by Step
$$\eqalign{
& \int_2^3 {\frac{{2x - 3}}{{\sqrt {4x - {x^2}} }}} dx \cr
& {\text{Completing the square}} \cr
& \int_2^3 {\frac{{2x - 3}}{{\sqrt {4x - {x^2}} }}} dx = \int_2^3 {\frac{{2x - 3}}{{\sqrt {4 - {{\left( {x - 2} \right)}^2}} }}} dx \cr
& {\text{Let }}u = x - 2,{\text{ }}x = u + 2,{\text{ }}dx = du,{\text{ }} \cr
& {\text{The new limits of integration are}} \cr
& x = 3 \Rightarrow u = 3 - 2 = 1 \cr
& x = 2 \Rightarrow u = 2 - 2 = 0 \cr
& \int_2^3 {\frac{{2x - 3}}{{\sqrt {4 - {{\left( {x - 2} \right)}^2}} }}} dx = \int_0^1 {\frac{{2\left( {u + 2} \right) - 3}}{{\sqrt {4 - {u^2}} }}} du \cr
& = \int_0^1 {\frac{{2u + 1}}{{\sqrt {4 - {u^2}} }}} du \cr
& = \int_0^1 {\left( {\frac{{2u}}{{\sqrt {4 - {u^2}} }} + \frac{1}{{\sqrt {4 - {u^2}} }}} \right)} du \cr
& {\text{Integrating}} \cr
& = \left[ { - 2\sqrt {4 - {u^2}} + {{\sin }^{ - 1}}\left( {\frac{u}{2}} \right)} \right]_0^1 \cr
& = \left[ { - 2\sqrt {4 - {1^2}} + {{\sin }^{ - 1}}\left( {\frac{1}{2}} \right)} \right] - \left[ { - 2\sqrt {4 - 0} + {{\tan }^{ - 1}}\left( {\frac{0}{2}} \right)} \right] \cr
& = - 2\sqrt 3 + \frac{1}{6}\pi + 4 \cr} $$
