Answer
$$\frac{{x + 1}}{2}\sqrt {3 - 2x - {x^2}} + \frac{4}{2}{\sin ^{ - 1}}\left( {\frac{{x + 1}}{2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {3 - 2x - {x^2}} } dx \cr
& {\text{completing the square for the radicand}} \cr
& = - \left( {{x^2} + 2x - 3} \right) \cr
& = - \left( {{x^2} + 2x - 3 + 1 - 1} \right) \cr
& = - \left[ {{{\left( {x + 1} \right)}^2} - 4} \right] \cr
& = 4 - {\left( {x + 1} \right)^2} \cr
& \cr
& \int {\sqrt {3 - 2x - {x^2}} } dx = \int {\sqrt {4 - {{\left( {x + 1} \right)}^2}} } dx \cr
& {\text{make }}u = x + 1,\,\,\,\,{\text{then}}\,\,\,\,\,\,du = dx \cr
& \int {\sqrt {4 - {{\left( {x + 1} \right)}^2}} } dx = \int {\sqrt {4 - {u^2}} } du \cr
& \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 74}} \cr
& \left( {74} \right):\,\,\,\,\,\,\int {\sqrt {{a^2} - {u^2}} } du = \frac{u}{2}\sqrt {{a^2} - {u^2}} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\frac{u}{a} + C \cr
& \int {\sqrt {4 - {u^2}} } du = \frac{u}{2}\sqrt {4 - {u^2}} + \frac{4}{2}{\sin ^{ - 1}}\frac{u}{2} + C \cr
& \cr
& {\text{write in terms of }}x{\text{; replace }}x + 1 {\text{ for }}u \cr
& = \frac{{x + 1}}{2}\sqrt {4 - {{\left( {x + 1} \right)}^2}} + \frac{4}{2}{\sin ^{ - 1}}\left( {\frac{{x + 1}}{2}} \right) + C \cr
& = \frac{{x + 1}}{2}\sqrt {3 - 2x - {x^2}} + \frac{4}{2}{\sin ^{ - 1}}\left( {\frac{{x + 1}}{2}} \right) + C \cr} $$
