Answer
$$\sqrt {{x^2} + 6x + 12} - 3\ln \left( {\sqrt {{x^2} + 6x + 12} + x + 3} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{\sqrt {{x^2} + 6x + 12} }}} dx \cr
& {\text{Completing the square}} \cr
& \int {\frac{x}{{\sqrt {{x^2} + 6x + 9 + 3} }}} dx = \int {\frac{x}{{\sqrt {\left( {{x^2} + 6x + 9} \right) + {{\left( {\sqrt 3 } \right)}^2}} }}} dx \cr
& = \int {\frac{x}{{\sqrt {{{\left( {x + 3} \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}} }}} dx \cr
& {\text{Let }}u = x + 3,{\text{ }}x = u - 3,{\text{ }}dx = du \cr
& \int {\frac{x}{{\sqrt {{{\left( {x + 3} \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}} }}} dx = \int {\frac{{u - 3}}{{\sqrt {{u^2} + {{\left( {\sqrt 3 } \right)}^2}} }}} du \cr
& {\text{Let }}u = \sqrt 3 \tan \theta ,{\text{ }}du = \sqrt 3 {\sec ^2}\theta d\theta \cr
& \int {\frac{{u - 3}}{{\sqrt {{u^2} + {{\left( {\sqrt 3 } \right)}^2}} }}} du = \int {\frac{{\sqrt 3 \tan \theta - 3}}{{\sqrt {3{{\tan }^2}\theta + 3} }}} \left( {\sqrt 3 {{\sec }^2}\theta d\theta } \right)d\theta \cr
& = \int {\frac{{\sqrt 3 \tan \theta - 3}}{{\sqrt {3\left( {{{\tan }^2}\theta + 1} \right)} }}} \left( {\sqrt 3 {{\sec }^2}\theta d\theta } \right)d\theta \cr
& {\text{Use the pythagorean identity }}{\tan ^2}\theta + 1 = {\sec ^2}\theta \cr
& = \int {\frac{{\sqrt 3 \tan \theta - 3}}{{\sqrt {3{{\sec }^2}\theta } }}} \left( {\sqrt 3 {{\sec }^2}\theta d\theta } \right)d\theta \cr
& = \int {\frac{{\sqrt 3 \tan \theta - 3}}{{\sqrt 3 \sec \theta }}} \left( {\sqrt 3 {{\sec }^2}\theta d\theta } \right)d\theta \cr
& = \int {\left( {\sqrt 3 \tan \theta - 3} \right)} \left( {\sec \theta d\theta } \right)d\theta \cr
& {\text{Distribute}} \cr
& = \int {\sqrt 3 \tan \theta \sec \theta d\theta } - 3\int {\sec \theta } d\theta \cr
& {\text{Integrating}} \cr
& = \sqrt 3 \sec \theta - 3\ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& {\text{We know that }}\tan \theta = \frac{u}{{\sqrt 3 }},{\text{ and sec}}\theta = \frac{{\sqrt {{u^2} + 3} }}{{\sqrt 3 }} \cr
& = \sqrt 3 \left( {\frac{{\sqrt {{u^2} + 3} }}{{\sqrt 3 }}} \right) - 3\ln \left| {\frac{{\sqrt {{u^2} + 3} }}{{\sqrt 3 }} + \frac{u}{{\sqrt 3 }}} \right| + C \cr
& = \sqrt {{u^2} + 3} - 3\ln \left| {\frac{{\sqrt {{u^2} + 3} + u}}{{\sqrt 3 }}} \right| + C \cr
& {\text{Substitute }}u = x + 3 \cr
& = \sqrt {{{\left( {x + 3} \right)}^2} + 3} - 3\ln \left| {\frac{{\sqrt {{{\left( {x + 3} \right)}^2} + 3} + x + 3}}{{\sqrt 3 }}} \right| + C \cr
& = \sqrt {{x^2} + 6x + 12} - 3\ln \left| {\frac{{\sqrt {{x^2} + 6x + 12} + x + 3}}{{\sqrt 3 }}} \right| + C \cr
& {\text{Using the properties of logarithms and combining constants}} \cr
& = \sqrt {{x^2} + 6x + 12} - 3\ln \left( {\sqrt {{x^2} + 6x + 12} + x + 3} \right) + C \cr} $$
