1. We need the domain of the composite function $f\circ g$, i.e. $f(g(x))$.

2. Given: $f(x)=\frac{2x+3}{2x+1}, \qquad \text{domain of } f = \mathbb{R}\setminus\left\{-\frac12\right\}$ $g(x)=\frac{|x|+1}{2x+5}, \qquad \text{domain of } g = \mathbb{R}\setminus\left\{-\frac52\right\}$

3. For $f(g(x))$ to be defined, two conditions are necessary:

   1. $g(x)$ must be defined.
   2. $g(x)$ must belong to the domain of $f$, i.e. $g(x)\neq -\frac12$

Step 1: Ensure $g(x)$ is defined

Since $g(x)=\frac{|x|+1}{2x+5},$ it is undefined when $2x+5=0 \implies x=-\frac52.$ So, we must exclude $x=-\frac52.$

Step 2: Ensure $g(x)\neq -\frac12$

We solve: $\frac{|x|+1}{2x+5}=-\frac12.$

Cross-multiplying: $2(|x|+1)=-(2x+5)$ $2|x|+2=-2x-5$ $2|x|+2x=-7.$

Now consider cases.

Case 1: $x\ge 0$

Then $|x|=x$, so $2x+2x=-7$ $4x=-7$ $x=-\frac74,$ which contradicts $x\ge 0$. So no solution here.

Case 2: $x<0$

Then $|x|=-x$, so $2(-x)+2x=-7$ $0=-7,$ which is impossible.

Hence, there is no real $x$ such that $g(x)=-\frac12.$

So no extra value needs to be excluded.

Step 3: Domain of $f\circ g$

Therefore, the only restriction comes from the domain of $g$: $\boxed{\mathbb{R}\setminus\left\{-\frac52\right\}}$

This corresponds to Option D.

Verification with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.