The value of the integral  $\underset{0}{\overset{\pi /2}{\int }}\frac{1+2\sin x}{{(2+\sin x)}^2}dx$ is equal to  

$I=\underset{0}{\overset{x}{\int }}\frac{1+2\sin x}{{(2+\sin x)}^2}dx$

Multiplying numerator and denominator by  ${\sec }^{2}x,$  we get 

$$\begin{gathered} I=\int \frac{{\sec }^{2}x+2\sec x\tan x}{{(2\sec x+\tan x)}^2}dxPut(2\sec x+\tan x)=t \\ \Rightarrow ({\sec }^2+2\sec x\tan x)dx=dt \\ I=\int \frac{dt}{t^2}=\frac{-1}{t}=\frac{-1}{2\sec x+\tan x}={\frac{-\cos x}{2+\sin x}]}_0^{\pi /2}=\frac{1}{2} \end{gathered}$$