Answer
$$\frac{1}{{\sqrt {10} }}{\tan ^{ - 1}}\left( {\sqrt {\frac{2}{5}} \left( {x + 1} \right)} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{2{x^2} + 4x + 7}}} \cr
& {\text{completing the square for }}2{x^2} + 4x + 7 \cr
& 2{x^2} + 4x + 7 = 2\left( {{x^2} + 2x + 1} \right) + 7 - 2 \cr
& 2{x^2} + 4x + 7 = 2{\left( {x + 1} \right)^2} + 5 \cr
& \int {\frac{{dx}}{{2{x^2} + 4x + 7}}} = \int {\frac{{dx}}{{2{{\left( {x + 1} \right)}^2} + 5}}} \cr
& = \frac{1}{2}\int {\frac{{dx}}{{{{\left( {x + 1} \right)}^2} + 5/2}}} \cr
& \cr
& {\text{Write in terms of }}\theta \cr
& {\text{substitute }}x + 1 = \left( {\sqrt {5/2} } \right)\tan \theta ,{\text{ }}\,\,\,dx = \left( {\sqrt {5/2} } \right){\sec ^2}\theta d\theta \cr
& = \frac{1}{2}\int {\frac{{dx}}{{{{\left( {x + 1} \right)}^2} + 5/2}}} = \frac{1}{2}\int {\frac{{\left( {\sqrt {5/2} } \right){{\sec }^2}\theta d\theta }}{{\left( {5/2} \right){{\tan }^2}\theta + 5/2}}} \cr
& = \frac{{\sqrt {5/2} }}{5}\int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\tan }^2}\theta + 1}}} \cr
& = \frac{1}{{\sqrt {10} }}\int {\frac{{{{\sec }^2}\theta d\theta }}{{{{\sec }^2}\theta }}} \cr
& = \frac{1}{{\sqrt {10} }}\int {d\theta } \cr
& {\text{Integrate}} \cr
& = \frac{1}{{\sqrt {10} }}\theta + C \cr
& \cr
& {\text{Where }}x + 1 = \left( {\sqrt {5/2} } \right)\tan \theta \to \theta = {\tan ^{ - 1}}\left( {\sqrt {\frac{2}{5}} \left( {x + 1} \right)} \right) \cr
& = \frac{1}{{\sqrt {10} }}{\tan ^{ - 1}}\left( {\sqrt {\frac{2}{5}} \left( {x + 1} \right)} \right) + C \cr} $$
