Answer
$$y = \frac{1}{2}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + \frac{{7\pi }}{8}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{1}{{4 + {x^2}}} \cr
& {\text{Separating variables}} \cr
& dy = \frac{1}{{4 + {x^2}}}dx \cr
& {\text{Integrating both sides}} \cr
& \int {dy} = \int {\frac{1}{{4 + {x^2}}}} dx \cr
& y = \frac{1}{2}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Using the initial condition }}y\left( 2 \right) = \pi \cr
& \pi = \frac{1}{2}{\tan ^{ - 1}}\left( {\frac{2}{2}} \right) + C \cr
& \pi = \frac{1}{2}\left( {\frac{\pi }{4}} \right) + C \cr
& \pi = \frac{\pi }{8} + C \cr
& C = \frac{{7\pi }}{8} \cr
& {\text{Substituting }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = \frac{1}{2}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + \frac{{7\pi }}{8} \cr} $$
