Answer
$${\sin ^{ - 1}}\left( {x + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt { - 2x - {x^2}} }}} \cr
& {\text{completing the square for }} - 2x - {x^2} \cr
& - 2x - {x^2} = - \left( {{x^2} + 2x + 1} \right) + 1 \cr
& - 2x - {x^2} = 1 - {\left( {x + 1} \right)^2} \cr
& {\text{then}}{\text{,}} \cr
& \int {\frac{{dx}}{{\sqrt {1 - {{\left( {x + 1} \right)}^2}} }}} \cr
& {\text{use the substitution method}} \cr
& u = x + 1,\,\,\,\,\,\,du = dx \cr
& \int {\frac{{dx}}{{\sqrt {1 - {{\left( {x + 1} \right)}^2}} }}} = \int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} \cr
& {\text{integrating}} \cr
& = {\sin ^{ - 1}}u + C \cr
& {\text{write in terms of }}x{\text{; replace }}u = x + 1 \cr
& = {\sin ^{ - 1}}\left( {x + 1} \right) + C \cr} $$
