The integral $$\int \frac{(2x-1)\cos\sqrt{(2x-1)^2+5}}{\sqrt{4x^2-4x+6}}dx$$ is equal to (where $$c$$ is a constant of integration):

Let $$t = (2x - 1)^2 + 5$$. Note that $$4x^2 - 4x + 6 = (2x-1)^2 + 5 = t$$, so $$\sqrt{4x^2 - 4x + 6} = \sqrt{t}$$.

Also, $$\frac{dt}{dx} = 2(2x-1) \cdot 2 = 4(2x-1)$$, so $$(2x-1)\,dx = \frac{dt}{4}$$.

The integral becomes $$\int \frac{(2x-1)\cos\sqrt{t}}{\sqrt{t}}\,dx = \int \frac{\cos\sqrt{t}}{\sqrt{t}} \cdot \frac{dt}{4}$$.

Now let $$u = \sqrt{t}$$, so $$du = \frac{dt}{2\sqrt{t}}$$, meaning $$\frac{dt}{\sqrt{t}} = 2\,du$$.

The integral becomes $$\int \frac{\cos u}{4} \cdot 2\,du = \frac{1}{2}\int \cos u\,du = \frac{1}{2}\sin u + c = \frac{1}{2}\sin\sqrt{t} + c$$.

Substituting back $$t = (2x-1)^2 + 5$$, we get $$\frac{1}{2}\sin\sqrt{(2x-1)^2 + 5} + c$$.

The answer is Option A: $$\frac{1}{2}\sin\sqrt{(2x-1)^2+5} + c$$.