Answer
$$\sqrt {28} - \sqrt {12} $$
Work Step by Step
$$\eqalign{
& \int_1^3 {\frac{{x + 2}}{{\sqrt {{x^2} + 4x + 7} }}} dx \cr
& u = {x^2} + 4x + 7,{\text{ then }}du = \left( {2x + 4} \right)dx,{\text{ }}\frac{1}{2}du = \left( {x + 2} \right)dx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = 3,{\text{ }}u = {\left( 3 \right)^2} + 4\left( 3 \right) + 7 = 28 \cr
& {\text{if }}x = 1,{\text{ }}u = {\left( 1 \right)^2} + 4\left( 1 \right) + 7 = 12 \cr
& {\text{so}} \cr
& \int_1^3 {\frac{{x + 2}}{{\sqrt {{x^2} + 4x + 7} }}} dx = \int_{12}^{28} {\frac{1}{{\sqrt u }}\left( {\frac{1}{2}du} \right)} \cr
& = \frac{1}{2}\int_{12}^{28} {{u^{ - 1/2}}du} \cr
& {\text{find the antiderivative }} \cr
& = \frac{1}{2}\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right)_{12}^{28} \cr
& = \left( {\sqrt u } \right)_{12}^{28} \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = \sqrt {28} - \sqrt {12} \cr} $$
