Let $ t = 2 + \cos x $. Then $ dt = -\sin x\, dx $. 

But the integral has no $\sin x$, so rewrite numerator: 

 $ 1 + 2\cos x = (2 + \cos x) - 1 = t - 1 $. 

 Now express $ dx $ using $ \sin^2 x = 1 - \cos^2 x $, but the standard trick is to differentiate: 

 $ \dfrac{d}{dx}\left(\dfrac{1}{2+\cos x}\right) = -\dfrac{-\sin x}{(2+\cos x)^2} = \dfrac{\sin x}{(2+\cos x)^2}. $ We 

use complementary substitution: 

 Let $ x = \frac{\pi}{2} - y $. 

 Then $\cos x = \sin y$ and $\sin x = \cos y$. 

 Integral becomes: 

 $ I = \int_{0}^{\pi/2} \frac{1 + 2\sin y}{(2 + \sin y)^2}\, dy. $ 

 Average the two forms: 

 $ I = \frac{1}{2}\int_{0}^{\pi/2} \left[ \frac{1 + 2\cos x}{(2 + \cos x)^2} + \frac{1 + 2\sin x}{(2 + \sin x)^2} \right] dx. $ 

 Now observe identity: 

 $ \frac{1 + 2\cos x}{(2 + \cos x)^2} + \frac{1 + 2\sin x}{(2 + \sin x)^2} = \frac{d}{dx}\left(\frac{\sin x - \cos x}{(2+\cos x)(2+\sin x)}\right). $ 

 Thus integral becomes a telescoping form and evaluates to: 

 $ I = \left[ \frac{\sin x - \cos x}{(2+\cos x)(2+\sin x)} \right]_{0}^{\pi/2}. $ 

 Now compute: 

 At $ x = \frac{\pi}{2}$: $ \sin x = 1,\;\cos x = 0 $ 

 Expression = $ \dfrac{1 - 0}{(2+0)(2+1)} = \dfrac{1}{6}. $ 

 At $ x = 0$: $ \sin 0 = 0,\;\cos 0 = 1 $ 

 Expression = $ \dfrac{0 - 1}{(2+1)(2+0)} = -\dfrac{1}{6}. $ 

 Therefore: $ I = \dfrac{1}{6} - (-\dfrac{1}{6}) = \dfrac{2}{6} = \dfrac{1}{3}. $ 

 Final Answer: $\dfrac{1}{3} $