Step 1: Rewrite the denominator
Use $\sin2x=2\sin x\cos x$ and $3=3\sin^{2}x+3\cos^{2}x$.
The denominator is $4\sin^{2}x+2\sin x\cos x+3\cos^{2}x$.
Step 2: Substitute t = tan x
Dividing by $\cos^{2}x$ and using $dt=\sec^{2}x\,dx$: $\int\frac{dt}{4t^{2}+2t+3}$.
Step 3: Standard form
$\int\frac{dt}{at^{2}+bt+c}$ with $4ac-b^{2}=48-4=44$ gives $\frac{2}{\sqrt{44}}\tan^{-1}\frac{8t+2}{\sqrt{44}}=\frac{1}{\sqrt{11}}\tan^{-1}\frac{4t+1}{\sqrt{11}}$.
Final Answer:
So the result is the same expression with root 11 and (4 tan x + 1).
\[ \boxed{\text{(D) }\dfrac{1}{\sqrt{11}}\tan ^{-1}\left(\dfrac{4\,\tan x+1}{\sqrt{11}}\right)+c} \]