Answer
$$\frac{{2\sqrt 3 }}{3}{\tan ^{ - 1}}\left( {\frac{{2\sqrt 3 \tan \left( {x/2} \right) + \sqrt 3 }}{3}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{2 + \sin x}}} \cr
& {\text{Use substitution }}\left( 5 \right){\text{ from the page 528}}: \cr
& \,\,u = \tan \left( {\frac{x}{2}} \right){\text{,}}\,\,\,\,\,\,dx = \frac{2}{{1 + {u^2}}}du,\,\,\,\,\,\,\sin x = \frac{{2u}}{{1 + {u^2}}}\,\,\,\,\,{\text{ and}}\,{\text{ }}\,\,\cos x = \frac{{1 - {u^2}}}{{1 + {u^2}}} \cr
& \cr
& \int {\frac{{dx}}{{2 + \sin x}}} = \int {\frac{{\frac{2}{{1 + {u^2}}}du}}{{2 + \frac{{2u}}{{1 + {u^2}}}}}} \cr
& {\text{simplify }} \cr
& = \int {\frac{{\frac{2}{{1 + {u^2}}}du}}{{\frac{{2 + 2{u^2} + 2u}}{{1 + {u^2}}}}}} \cr
& = \int {\frac{2}{{2 + 2{u^2} + 2u}}} du \cr
& = \int {\frac{1}{{{u^2} + u + 1}}} du \cr
& \cr
& {\text{Completing the square for }}{u^2} + u + 1 \cr
& = \int {\frac{1}{{{u^2} + u + \frac{1}{4} + \frac{3}{4}}}} du \cr
& = \int {\frac{1}{{{{\left( {u + \frac{1}{2}} \right)}^2} + \frac{3}{4}}}} du \cr
& {\text{Integrate by the formula }}\int {\frac{{du}}{{{x^2} + {a^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + C \cr
& = \frac{1}{{\left( {\sqrt 3 /2} \right)}}{\tan ^{ - 1}}\left( {\frac{{u + 1/2}}{{\sqrt 3 /2}}} \right) + C \cr
& = \frac{{2\sqrt 3 }}{3}{\tan ^{ - 1}}\left( {\frac{{2\sqrt 3 \left( {u + 1/2} \right)}}{3}} \right) + C \cr
& = \frac{{2\sqrt 3 }}{3}{\tan ^{ - 1}}\left( {\frac{{2\sqrt 3 u + \sqrt 3 }}{3}} \right) + C \cr
& \cr
& {\text{write in terms of }}x{\text{; substitute }}u = \tan \left( {\frac{x}{2}} \right) \cr
& = \frac{{2\sqrt 3 }}{3}{\tan ^{ - 1}}\left( {\frac{{2\sqrt 3 \tan \left( {x/2} \right) + \sqrt 3 }}{3}} \right) + C \cr} $$
