Answer
$${\sin ^{ - 1}}\left( {x + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt { - 2x - {x^2}} }}} \cr
& {\text{Complete the square to the radicand }} - 2x - {x^2} \cr
& - 2x - {x^2} = - \left( {{x^2} + 2x + 1} \right) + 1 \cr
& 2x - {x^2} = 1 - {\left( {x + 1} \right)^2} \cr
& \cr
& \int {\frac{{dx}}{{\sqrt { - 2x - {x^2}} }}} = \int {\frac{{dx}}{{\sqrt {1 - {{\left( {x + 1} \right)}^2}} }}} \cr
& \cr
& {\text{Let }}u = x + 1,\,\,\,\,du = dx \cr
& \,\,\,\,\,\,\,\,\int {\frac{{dx}}{{\sqrt {1 - {{\left( {x + 1} \right)}^2}} }}} = \int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} \cr
& \cr
& {\text{Recall that }}\int {\frac{{dx}}{{\sqrt {1 - {x^2}} }} = {{\sin }^{ - 1}}x + C} \cr
& \cr
& \int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} = {\sin ^{ - 1}}u + C \cr
& \cr
& {\text{Write in terms of }}x,{\text{ replacing }}x + 1{\text{ for }}u \cr
& = {\sin ^{ - 1}}\left( {x + 1} \right) + C \cr} $$
