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The method that works best is Weierstrass substitution (hope the spelling's correct). It starts by defining $t=\tan(x/2)$ . Then we have $\cos x=1-2\sin^2(x/2)=1-\dfrac{2t^2}{\sec^2(x/2)}=1-\dfrac{2t^2}{1+t^2}=\dfrac{1-t^2}{1+t^2}.$ Furthermore we have $\dfrac{\text d t}{\text d x}=\dfrac 1 2\sec^2(x/2)=\dfrac{1+t^2}{2}\implies \text d x=\dfrac{2~\text d t}{1+t^2}.$ Subbing all these and simplifying yields $\int \dfrac{\text d x}{2+\cos x}=\int \dfrac{2~\text d t}{3+t^2}.$ Now let $t=\sqrt 3 \tan z$ so that $\text d t=\sqrt 3 \sec^2 z~\text d z$ . Subbing these gives $\int \dfrac{2~\text d t}{3+t^2}=2\int \dfrac{\sqrt 3\sec^2 z~\text d z }{3+3\tan^2 z}=2\int\dfrac{\sqrt 3 \sec^2 z~\text d z}{3\sec^2 z}=2\int\dfrac{\text d z}{\sqrt 3}=\dfrac{2z}{\sqrt3}+\text C.$ Now subbing back $z=\arctan\left(\dfrac{t}{\sqrt 3}\right)=\arctan\left[\dfrac{\tan(x/2)}{\sqrt 3}\right]$ gives $\int \dfrac{\text d x}{2+\cos x}=\dfrac{2}{\sqrt 3}\arctan\left[\dfrac{\tan(x/2)}{\sqrt 3}\right]+\text C.$ So plugging in the values we get $\pi\sqrt 3/9$ . Hence the answer is $\boxed{13}$ .

$I= \displaystyle\int\frac{1}{2+cosx}\,dx$ $I=\displaystyle\int\frac{1}{1+(1+cosx)}\,dx$ $I=\displaystyle\int\frac{1}{sin^2(x/2) +cos^2(x/2)+2cos^2(x/2)}\,dx$ \ $I= \displaystyle\int\frac{1}{sin^2 (x/2)+3cos^2(x/2)}\,dx$ Multiplying and dividing by $sec^2(x/2)$ in above expression we get $I = \displaystyle\int\frac{sec^2(x/2)}{tan^2(x/2)+3}\,dx$ Now substituting and differentiating $tan(x/2) = y \Rightarrow sec^2(x/2)\,dx = 2\,dy$ $I=\displaystyle\int\frac{2}{y^2+3}\,dy$ $=2\displaystyle \int\frac{\,dy}{y^2+(\sqrt{3})^2}$ Applying upper and lower limits in above integration we get : $I =2\displaystyle\int_{0}^{\pi/2}\frac{\,dy}{y^2+(\sqrt{3})^2}$ $= \frac{2}{\sqrt 3}\left[tan^{-1}\left(\frac{y}{\sqrt 3}\right)\right]_{0}^{\pi/2}$ Replugging $y = tan(x/2)$ we get as follows; $I =\frac{2}{\sqrt 3}\left[tan^{-1}\left(\frac{tan(x/2)}{\sqrt 3}\right)\right]_{0}^{\pi/2}$

Setting the upper limit and lower limit we get as $I = \frac{2}{\sqrt 3}\left[tan^{-1}\left(\frac{1}{\sqrt 3}\right)-tan^{-1}\left(\frac{0}{\sqrt 3}\right)\right]$ $I = \frac{\pi\sqrt 3}{9}$

Therefore $A+B+C= 1+9+3\implies 13$ .

The problem can be solved quickly by putting $cosx = \frac{1-tan^2(x/2)}{1+tan^2(x/2)}$ that directly makes us jump to $\displaystyle\int\frac{sec^2(x/2)}{tan^2(x/2)+3}\,dx$ .