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