Answer
$$ - x - 2\sqrt {x + 1} + {C_1}$$
Work Step by Step
$$\eqalign{
& \int {\frac{{ - x}}{{\left( {x + 1} \right) - \sqrt {x + 1} }}} dx \cr
& {\text{Integrate by substitution}} \cr
& u = x + 1,{\text{ }}x = u - 1,{\text{ }}dx = du \cr
& {\text{Substituting}} \cr
& \int {\frac{{ - x}}{{\left( {x + 1} \right) - \sqrt {x + 1} }}} dx = \int {\frac{{ - \left( {u - 1} \right)}}{{u - \sqrt u }}} du \cr
& = - \int {\frac{{u - 1}}{{u - \sqrt u }}} du \cr
& {\text{Rationalizing the denominator}} \cr
& = - \int {\frac{{u - 1}}{{u - \sqrt u }} \times \left( {\frac{{u + \sqrt u }}{{u + \sqrt u }}} \right)} du \cr
& = - \int {\frac{{\left( {u - 1} \right)\left( {u + \sqrt u } \right)}}{{{u^2} - {{\left( {\sqrt u } \right)}^2}}}} du \cr
& = - \int {\frac{{\left( {u - 1} \right)\left( {u + \sqrt u } \right)}}{{{u^2} - u}}} du \cr
& {\text{Factor the numerator and denominator}} \cr
& = - \int {\frac{{\sqrt u \left( {u - 1} \right)\left( {\sqrt u + 1} \right)}}{{u\left( {u - 1} \right)}}} du \cr
& {\text{Divide}} \cr
& = - \int {\frac{{\sqrt u + 1}}{{\sqrt u }}} du \cr
& = - \int {\left( {1 + {u^{ - 1/2}}} \right)} du \cr
& {\text{Integrating}} \cr
& = - \left( {u + \frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr
& = - u - 2\sqrt u + C \cr
& {\text{Write in terms of }}x \cr
& = - \left( {x + 1} \right) - 2\sqrt {x + 1} + C \cr
& = - x - 1 - 2\sqrt {x + 1} + C \cr
& {\text{Combine constants}} \cr
& = - x - 2\sqrt {x + 1} + {C_1} \cr} $$
