The given integral can be rewritten as,

$I=\int \frac{\sin x+\cos x}{3+\sin 2x}dx+\int \frac{\sin x-\cos x}{3+\sin 2x}dx$
Let $\sin x-\cos x=t$ and $\sin x+\cos x=u$ in $1^{\mathrm{s}\mathrm{t}}$ and $2^{\mathrm{n}\mathrm{d}}$ integral respectively.
$\therefore I=\int \frac{dt}{4-t^2}-\int \frac{du}{2+u^2}$
$=\frac{1}{4}\ln \left|\frac{2+t}{2-t}\right|-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{u}{\sqrt{2}}\right)+C$
Putting values of $t$ and $u$, we get,

$I=\frac{1}{4}\ln \left|\frac{2+\sin x-\cos x}{2-\sin x+\cos x}\right|-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right)+C$