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