$\text{ Let }g\left(x\right)=\int \frac{1+2\cos x}{(\cos x+2{)}^2}dx\text{ and }g\left(0\right)=0\text{ . Then the value of }g(3\pi /2)\text{ is }$

$\begin{array}{l}g\left(x\right)=\int \frac{1+2\cos x}{(\cos x+2{)}^2}dx\\ =\int \frac{{\mathrm{cosec}}^{2}x+2\cot x\mathrm{cosec}x}{(\cot x+2\mathrm{cosec}x{)}^2}dx\\ \text{ (Dividing numerator and denominator by }{\sin }^2\mathrm{x}\text{ ) }\end{array}$

$\begin{array}{l}=\frac{1}{\cot x+2\mathrm{cosec}x}+C\\ g\left(x\right)=\frac{\sin x}{\cos x+2}+C\\ \text{ Since }g\left(0\right)=0\text{ , we get that }\mathrm{C}=0\\ \text{ So, }g\left(x\right)=\frac{\sin x}{\cos x+2}\\ \therefore \mathrm{g}(3\pi /2)=-\frac{1}{2}\end{array}$