1. Rewrite the denominator

Given $I=\int \frac{(2x-1)\cos\sqrt{(2x-1)^2+5}}{\sqrt{4x^2-4x+6}}\,dx$

Notice that $4x^2-4x+6=(2x-1)^2+5$

because $(2x-1)^2+5=4x^2-4x+1+5=4x^2-4x+6.$

So the integral becomes $I=\int \frac{(2x-1)\cos\sqrt{(2x-1)^2+5}}{\sqrt{(2x-1)^2+5}}\,dx.$

2. Use substitution

Let $t=\sqrt{(2x-1)^2+5}.$ Then $t^2=(2x-1)^2+5.$

Differentiate both sides: $2t\,\frac{dt}{dx}=2(2x-1)\cdot 2=4(2x-1).$

Hence $t\,dt=2(2x-1)\,dx$ so $\frac{(2x-1)}{t}\,dx=\frac{1}{2}\,dt.$ Since $t=\sqrt{(2x-1)^2+5}$, the integral becomes $I=\int \cos t\left(\frac{(2x-1)}{t}dx\right)=\int \cos t\cdot \frac12\,dt.$

Thus $I=\frac12\int \cos t\,dt=\frac12\sin t+c.$

3. Substitute back

$I=\frac12\sin\sqrt{(2x-1)^2+5}+c.$

4. Match with the options

This is exactly Option A: $\boxed{\frac12\sin\sqrt{(2x-1)^2+5}+c}$

5. Verification by differentiation

Differentiate $F(x)=\frac12\sin\sqrt{(2x-1)^2+5}.$

Then $F'(x)=\frac12\cos\sqrt{(2x-1)^2+5}\cdot \frac{1}{2\sqrt{(2x-1)^2+5}}\cdot 4(2x-1)$ $=\frac{(2x-1)\cos\sqrt{(2x-1)^2+5}}{\sqrt{(2x-1)^2+5}}$ $=\frac{(2x-1)\cos\sqrt{(2x-1)^2+5}}{\sqrt{4x^2-4x+6}},$ which matches the integrand.

Therefore, the answer is correct.