For $$x^2 \neq n\pi + 1$$, $$n \in N$$ (the set of natural numbers), the integral $$\int x\sqrt{\frac{2\sin(x^2-1) - \sin 2(x^2-1)}{2\sin(x^2-1) + \sin 2(x^2-1)}} \; dx$$ is equal to (where c is a constant of integration):

We need to evaluate $$\int x\sqrt{\frac{2\sin(x^2-1) - \sin 2(x^2-1)}{2\sin(x^2-1) + \sin 2(x^2-1)}} \; dx$$

Let $$\theta = x^2 - 1$$. We simplify the expression inside the square root.

Using the identity $$\sin 2\theta = 2\sin\theta\cos\theta$$, the numerator becomes:

$$2\sin\theta - \sin 2\theta = 2\sin\theta - 2\sin\theta\cos\theta = 2\sin\theta(1 - \cos\theta)$$

The denominator becomes:

$$2\sin\theta + \sin 2\theta = 2\sin\theta + 2\sin\theta\cos\theta = 2\sin\theta(1 + \cos\theta)$$

So the ratio inside the square root is:

$$\frac{2\sin\theta(1-\cos\theta)}{2\sin\theta(1+\cos\theta)} = \frac{1-\cos\theta}{1+\cos\theta}$$

Using the half-angle identities $$1 - \cos\theta = 2\sin^2\frac{\theta}{2}$$ and $$1 + \cos\theta = 2\cos^2\frac{\theta}{2}$$:

$$\frac{1-\cos\theta}{1+\cos\theta} = \frac{2\sin^2(\theta/2)}{2\cos^2(\theta/2)} = \tan^2\frac{\theta}{2}$$

Therefore:

$$\sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \left|\tan\frac{\theta}{2}\right| = \tan\frac{\theta}{2}$$

(taking the positive value since $$x^2 - 1 \neq n\pi$$)

The integral becomes:

$$\int x \tan\frac{x^2-1}{2} \; dx$$

Let $$u = \frac{x^2-1}{2}$$, then $$du = x \; dx$$.

$$= \int \tan u \; du = \int \frac{\sin u}{\cos u} \; du = -\ln|\cos u| + c = \ln|\sec u| + c$$

Substituting back $$u = \frac{x^2-1}{2}$$:

$$= \ln\left|\sec\frac{x^2-1}{2}\right| + c$$

Now, $$\ln\left|\sec\frac{x^2-1}{2}\right| = \frac{1}{2}\ln\left|\sec^2\frac{x^2-1}{2}\right|$$

(since $$\ln|a| = \frac{1}{2}\ln|a^2|$$)

This matches Option C: $$\frac{1}{2}\log_e \left|\sec^2\frac{x^2-1}{2}\right| + c$$

The correct answer is Option C.