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