Answer
$${\sin ^{ - 1}}\left( {\frac{{x - 1}}{2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt {3 + 2x - {x^2}} }}} \cr
& {\text{completing the square for }}3 + 2x - {x^2} \cr
& 3 + 2x - {x^2} = - \left( {{x^2} - 2x + 1} \right) + 4 \cr
& 3 + 2x - {x^2} = 4 - {\left( {x - 1} \right)^2} \cr
& \int {\frac{{dx}}{{\sqrt {3 + 2x - {x^2}} }}} = \int {\frac{{dx}}{{\sqrt {4 - {{\left( {x - 1} \right)}^2}} }}} \cr
& \cr
& {\text{Write in terms of }}\theta \cr
& {\text{substitute }}x = 2\sin \theta + 1,{\text{ }}\,\,\,dx = 2\cos \theta d\theta \cr
& \int {\frac{{dx}}{{\sqrt {4 - {{\left( {x - 1} \right)}^2}} }}} = \int {\frac{{2\cos \theta d\theta }}{{\sqrt {4 - {{\left( {2\sin \theta + 1 - 1} \right)}^2}} }}} \cr
& = \int {\frac{{2\cos \theta d\theta }}{{\sqrt {4 - 4{{\sin }^2}\theta } }}} \cr
& = \int {\frac{{2\cos \theta d\theta }}{{2\sqrt {1 - {{\sin }^2}\theta } }}} \cr
& = \int {\frac{{\cos \theta d\theta }}{{\sqrt {{{\cos }^2}\theta } }}} \cr
& = \int {d\theta } \cr
& {\text{Integrate}} \cr
& = \theta + C \cr
& \cr
& {\text{Where }}x = 2\sin \theta + 1 \to \theta = {\sin ^{ - 1}}\left( {\frac{{x - 1}}{2}} \right) \cr
& = {\sin ^{ - 1}}\left( {\frac{{x - 1}}{2}} \right) + C \cr} $$
