The integral $$\displaystyle\int \sqrt{1 + 2\cot x(\csc x + \cot x)}\,dx$$, $$\left(0 \lt x \lt \frac{\pi}{2}\right)$$ is equal to:

We begin with the given integral

$$I=\displaystyle\int \sqrt{1+2\cot x\;(\csc x+\cot x)}\,dx,$$

where $$0\lt x\lt \dfrac{\pi}{2}.$$ Inside the square-root we have the expression

$$1+2\cot x\;(\csc x+\cot x).$$

Let us expand it term by term. First write every trigonometric function in terms of $$\sin x$$ and $$\cos x$$:

$$\cot x=\dfrac{\cos x}{\sin x},$$, $$\csc x=\dfrac{1}{\sin x}.$$

Hence

$$\begin{aligned} 2\cot x\;\csc x &=2\left(\dfrac{\cos x}{\sin x}\right)\left(\dfrac{1}{\sin x}\right)=\dfrac{2\cos x}{\sin^2 x},\\[6pt] 2\cot^2 x &=2\left(\dfrac{\cos x}{\sin x}\right)^2=\dfrac{2\cos^2 x}{\sin^2 x}. \end{aligned}$$

So we can write

$$1+2\cot x\;(\csc x+\cot x)=1+2\cot x\csc x+2\cot^2 x.$$

Now recall the Pythagorean identity $$\csc^2 x=1+\cot^2 x.$$ Using this, compute the square of the sum $$\bigl(\cot x+\csc x\bigr)^2$$:

$$\begin{aligned} (\cot x+\csc x)^2 &=\cot^2 x+\csc^2 x+2\cot x\csc x\\ &=\cot^2 x+\bigl(1+\cot^2 x\bigr)+2\cot x\csc x\\ &=1+2\cot^2 x+2\cot x\csc x. \end{aligned}$$

We see that this is exactly the same expression that appears under the square-root. Therefore

$$1+2\cot x\;(\csc x+\cot x)=\bigl(\cot x+\csc x\bigr)^2.$$

Because $$0\lt x\lt \dfrac{\pi}{2},$$ both $$\sin x$$ and $$\cos x$$ are positive, so $$\cot x+\csc x\gt 0$$ in this interval. Consequently,

$$\sqrt{1+2\cot x\;(\csc x+\cot x)}=\cot x+\csc x.$$

The integral now simplifies to

$$I=\displaystyle\int\bigl(\cot x+\csc x\bigr)\,dx.$$

We integrate the two terms separately. The standard integrals we need are:

$$\int\cot x\,dx=\ln|\sin x|+C_1,$$

and

$$\int\csc x\,dx=\ln\left|\tan\dfrac{x}{2}\right|+C_2.$$

Using these, we obtain

$$\begin{aligned} I &=\bigl[\ln|\sin x|\bigr]+\bigl[\ln\left|\tan\dfrac{x}{2}\right|\bigr]+C\\[6pt] &=\ln\left|\sin x\;\tan\dfrac{x}{2}\right|+C. \end{aligned}$$

To combine the logarithms further, recall the half-angle relation

$$\tan\dfrac{x}{2}=\dfrac{\sin x}{1+\cos x}.$$

So

$$\sin x\;\tan\dfrac{x}{2} =\sin x\;\dfrac{\sin x}{1+\cos x} =\dfrac{\sin^2 x}{1+\cos x}.$$

Next use the double-angle identities

$$\sin x=2\sin\dfrac{x}{2}\cos\dfrac{x}{2},$$, $$1+\cos x=2\cos^2\dfrac{x}{2}.$$

Substituting these, we have

$$\dfrac{\sin^2 x}{1+\cos x} =\dfrac{\bigl(2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\bigr)^2}{2\cos^2\dfrac{x}{2}} =\dfrac{4\sin^2\dfrac{x}{2}\cos^2\dfrac{x}{2}}{2\cos^2\dfrac{x}{2}} =2\sin^2\dfrac{x}{2}.$$

Therefore

$$I=\ln\left|2\sin^2\dfrac{x}{2}\right|+C.$$

Since $$\ln(2)$$ is merely a constant, it can be absorbed into the overall constant of integration. Thus we may write

$$I=2\ln\left|\sin\dfrac{x}{2}\right|+C,$$

where $$C$$ is an arbitrary constant.

Hence, the correct answer is Option A.