Step 1: Expand the expression inside the square root.
We have $1 + 2\cot x(\cot x + \cosec x) = 1 + 2\cot^2 x + 2\cot x \cosec x$.
Step 2: Recognize a perfect square using the Pythagorean identity.
Using $\cosec^2 x = 1 + \cot^2 x$, substitute $1 = \cosec^2 x - \cot^2 x$: \[\cosec^2 x - \cot^2 x + 2\cot^2 x + 2\cot x \cosec x = \cosec^2 x + \cot^2 x + 2\cot x \cosec x = (\cosec x + \cot x)^2\] So $\sqrt{1 + 2\cot x(\cot x + \cosec x)} = \cosec x + \cot x$.
Step 3: Use the half-angle identity to simplify.
Using $1 + \cos x = 2\cos^2\frac{x}{2}$ and $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$: \[\cosec x + \cot x = \frac{1+\cos x}{\sin x} = \frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \cot\frac{x}{2}\]
Step 4: Rewrite the integral.
The integral becomes $\int \cot\frac{x}{2}\, dx$.
Step 5: Evaluate using substitution.
Let $u = \frac{x}{2}$, $dx = 2\,du$: \[\int \cot\frac{x}{2}\, dx = 2\int \cot u\, du = 2\ln|\sin u| + C\]
Step 6: Substitute back and state the final answer.
Replacing $u = \frac{x}{2}$: \[\boxed{2\log\!\left(\sin\dfrac{x}{2}\right) + C}\]