$$I = \int {{{dx} \over {\cos x + \sqrt 3 \sin x}}} $$

$$ \Rightarrow I = \int {{{dx} \over {2\left[ {{1 \over 2}\cos x + {{\sqrt 3 } \over 2}\sin x} \right]}}} $$

$$ = {1 \over 2}\int {{{dx} \over {\left[ {\sin {\pi \over 6}\cos x + \cos {\pi \over 6}\sin x} \right]}}} $$

$$ = {1 \over 2}.\int {{{dx} \over {\sin \left( {x + {\pi \over 6}} \right)}}} $$

$$ \Rightarrow I = {1 \over 2}.\int {\cos ec\left( {x + {\pi \over 6}} \right)dx} $$

But we know that

$$\int {\cos ec\,x\,dx} = \log \left| {\left( {\tan x/2} \right)} \right| + C$$

$$\therefore$$ $$I = {1 \over 2}.\log \,\tan \left( {{x \over 2} + {\pi \over {12}}} \right) + C$$