Answer
$$\sqrt {{x^2} + 2x + 2} + 2\ln \left( {x + 1 + \sqrt {{x^2} + 2x + 2} } \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{x + 3}}{{\sqrt {{x^2} + 2x + 2} }}} dx \cr
& {\text{completing the square for }}{x^2} + 2x + 2 \cr
& = \int {\frac{{x + 3}}{{\sqrt {{x^2} + 2x + 1 + 1} }}} dx \cr
& = \int {\frac{{x + 3}}{{\sqrt {{{\left( {x + 1} \right)}^2} + 1} }}} dx \cr
& {\text{Write the integrand in terms of }}u \cr
& u = x + 1,\,\,\,du = dx \cr
& {\text{Then}}{\text{,}} \cr
& = \int {\frac{{u - 1 + 3}}{{\sqrt {{u^2} + 1} }}} du \cr
& = \int {\frac{{u + 2}}{{\sqrt {{u^2} + 1} }}} du \cr
& {\text{distribute the numerator}} \cr
& = \int {\frac{u}{{\sqrt {{u^2} + 1} }}} du + \int {\frac{2}{{\sqrt {{u^2} + 1} }}} du \cr
& = \frac{1}{2}\int {\frac{{2u}}{{\sqrt {{u^2} + 1} }}} du + \int {\frac{2}{{\sqrt {{u^2} + 1} }}} du \cr
& \cr
& {\text{Integrate }}\int {\frac{{2u}}{{\sqrt {{u^2} + 1} }}} du{\text{ by the power rule and }}\int {\frac{2}{{\sqrt {{u^2} + 1} }}} du{\text{ by the formulas }} \cr
& {\text{on the endapaper}}{\text{, formula }}\left( {75} \right){\text{ }}\int {\frac{{du}}{{\sqrt {{u^2} + {a^2}} }}} = \ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C \cr
& {\text{then}}{\text{,}} \cr
& = \sqrt {{u^2} + 1} + 2\ln \left( {u + \sqrt {{u^2} + 1} } \right) + C \cr
& \cr
& {\text{Write the integrand in terms of }}x,\,\,\,\,{\text{substitute }}x + 1{\text{ for }}u \cr
& = \sqrt {{{\left( {x + 1} \right)}^2} + 1} + 2\ln \left( {x + 1 + \sqrt {{{\left( {x + 1} \right)}^2} + 1} } \right) + C \cr
& = \sqrt {{x^2} + 2x + 2} + 2\ln \left( {x + 1 + \sqrt {{x^2} + 2x + 2} } \right) + C \cr} $$
