Put $\sin x = \frac{2 \tan x/2}{1 + \tan ^2 x/2}$ and $\cos x = \frac{1-\tan^2 x/2}{1+ \tan^2 x/2}$
$\therefore I = \begin{align*}\int \frac{dx}{ 3+ 2 . (\frac{2 \tan x/2}{1+tan^2 x/2})+ (\frac{1-\tan^2 x/2}{1+\tan^2 x/2} )} \end{align*} $
$= \begin{align*}\int \frac{ (1+ \tan^2 x/2) dx}{3 +3 \tan^2 x/2 + 4 \tan x/2 + 1-\tan^2 x/2} \end{align*} $
$= \begin{align*}\int \frac{ \sec^2 {x/2}\; dx}{2 \tan^2 x/2 + 4 \tan {x/2} + 4} dx \end{align*}$
$=\frac{1}{2} \begin{align*}\int \frac{ \sec^2 {x/2} \;dx}{ \tan^2 x/2 + 2 \tan {x/2} + 2} dx \end{align*}$
put $\tan x/2 = t \; \; \; \ ; \ \therefore \frac{1}{2} \sec^2 \; x/2 \;dx = dt $
$\therefore I = \frac{2}{2} \begin{align*}\int \frac{dt} { t^2+ 2t+2 } \end{align*}$
$= \begin{align*}\int \frac{dt}{(t+1)^2 +1} \end{align*}$
$= \tan^{-1} (1+t) + C$
$= \tan^{-1} (1+t) + C$
$= \tan^{-1} (1+ \tan^{-1} \frac{x}{2}) + C$
$\implies I = \tan^{-1} (1+ \tan^{-1} \frac{x}{2}) + C$