Answer
$ - \cos \left( {\sqrt {1 + {x^2}} } \right) + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{x\sin \left( {\sqrt {1 + {x^2}} } \right)}}{{\sqrt {1 + {x^2}} }}} dx \cr
& = \int {\sin \left( {\sqrt {1 + {x^2}} } \right)\left( {\frac{x}{{\sqrt {1 + {x^2}} }}} \right)} dx \cr
& {\text{Let }}u = \sqrt {1 + {x^2}} ,{\text{ }}du = \frac{x}{{\sqrt {1 + {x^2}} }}dx \cr
& {\text{Applying the substitution}}{\text{, we obtain}} \cr
& = \int {\sin udu} \cr
& {\text{Integrating}} \cr
& = - \cos u + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}\sqrt {1 + {x^2}} {\text{ for }}u \cr
& = - \cos \left( {\sqrt {1 + {x^2}} } \right) + C \cr} $$
