Answer
$$y = \frac{1}{{\sqrt {12} }}\arctan \left( {\frac{x}{{\sqrt {12} }}} \right) + 2 - \frac{{\sqrt 3 }}{6}\arctan \left( {\frac{{2\sqrt 3 }}{3}} \right)$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{1}{{12 + {x^2}}} \cr
& {\text{Separate the variables}} \cr
& dy = \frac{1}{{12 + {x^2}}}dx \cr
& {\text{Integrate both sides}} \cr
& \int {dy} = \int {\frac{1}{{12 + {x^2}}}} dx \cr
& {\text{By Theorem 5}}{\text{.17 }}\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}\arctan \left( {\frac{u}{a}} \right) + C} \cr
& y = \frac{1}{{\sqrt {12} }}\arctan \left( {\frac{x}{{\sqrt {12} }}} \right) + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}y\left( 4 \right) = 2 \cr
& 2 = \frac{1}{{\sqrt {12} }}\arctan \left( {\frac{4}{{\sqrt {12} }}} \right) + C \cr
& 2 = \frac{{\sqrt 3 }}{6}\arctan \left( {\frac{{2\sqrt 3 }}{3}} \right) + C \cr
& C = 2 - \frac{{\sqrt 3 }}{6}\arctan \left( {\frac{{2\sqrt 3 }}{3}} \right) \cr
& {\text{Substituting }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = \frac{1}{{\sqrt {12} }}\arctan \left( {\frac{x}{{\sqrt {12} }}} \right) + 2 - \frac{{\sqrt 3 }}{6}\arctan \left( {\frac{{2\sqrt 3 }}{3}} \right) \cr} $$
