The value of $$\int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \left(\left(\frac{x+1}{x-1}\right)^2 + \left(\frac{x-1}{x+1}\right)^2 - 2\right)^{\frac{1}{2}} dx$$ is:

Let the given integral be $$I$$.

$$I = \int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \left(\left(\frac{x+1}{x-1}\right)^2 + \left(\frac{x-1}{x+1}\right)^2 - 2\right)^{\frac{1}{2}} dx$$

The algebraic expression inside the square root can be simplified using the identity $$(a-b)^2 = a^2 + b^2 - 2ab$$.

Let $$a = \frac{x+1}{x-1}$$ and $$b = \frac{x-1}{x+1}$$.

Since $$ab = 1$$, the expression matches the identity perfectly.

$$\left(\left(\frac{x+1}{x-1}\right)^2 + \left(\frac{x-1}{x+1}\right)^2 - 2\right) = \left(\frac{x+1}{x-1} - \frac{x-1}{x+1}\right)^2$$

Substituting this back into the integral gives:

$$I = \int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \sqrt{\left(\frac{x+1}{x-1} - \frac{x-1}{x+1}\right)^2} dx$$

Using the property $$\sqrt{u^2} = \left|u\right|$$, we can rewrite the integral using absolute value.

$$I = \int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \left| \frac{x+1}{x-1} - \frac{x-1}{x+1} \right| dx$$

Simplify the fraction inside the absolute value.

$$\frac{x+1}{x-1} - \frac{x-1}{x+1} = \frac{(x+1)^2 - (x-1)^2}{(x-1)(x+1)}$$

$$\frac{(x+1)^2 - (x-1)^2}{(x-1)(x+1)} = \frac{4x}{x^2-1}$$

Now substitute this back into the integral.

$$I = \int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \left| \frac{4x}{x^2-1} \right| dx$$

Let the integrand be $$f(x) = \left| \frac{4x}{x^2-1} \right| $$.

Notice that replacing $$x$$ with $$-x$$ gives:

$$f(-x) = \left| \frac{4(-x)}{(-x)^2-1} \right| = \left| \frac{-4x}{x^2-1} \right| = \left| \frac{4x}{x^2-1} \right| = f(x)$$

This means the integrand is an even function.

Using the property of definite integrals for even functions, we can rewrite the limits.

$$I = 2 \int_{0}^{\frac{1}{\sqrt{2}}} \left| \frac{4x}{x^2-1} \right| dx$$

In the interval from $$x = 0$$ to $$x = \frac{1}{\sqrt{2}}$$, the value of $$x$$ is positive, so $$4x \ge 0$$.

The denominator satisfies $$x^2 \le \frac{1}{2}$$, which means $$x^2-1 < 0$$.

Since the numerator is positive and the denominator is negative, the entire fraction is negative.

To remove the absolute value, we multiply the expression by $$-1$$.

$$\left| \frac{4x}{x^2-1} \right| = -\frac{4x}{x^2-1} = \frac{4x}{1-x^2}$$

Substitute this back into the integral.

$$I = 2 \int_{0}^{\frac{1}{\sqrt{2}}} \frac{4x}{1-x^2} dx$$

$$I = 8 \int_{0}^{\frac{1}{\sqrt{2}}} \frac{x}{1-x^2} dx$$

To solve this integral, use the substitution method.

Let $$t = 1-x^2$$.

Differentiating both sides gives $$dt = -2x dx$$, which means $$x dx = -\frac{dt}{2}$$.

Now determine the new limits of integration.

When $$x = 0$$, the value of $$t$$ is $$t = 1-0^2 = 1$$.

When $$x = \frac{1}{\sqrt{2}}$$, the value of $$t$$ is $$t = 1-\left(\frac{1}{\sqrt{2}}\right)^2 = 1-\frac{1}{2} = \frac{1}{2}$$.

Substitute these values into the integral.

$$I = 8 \int_{1}^{\frac{1}{2}} \frac{1}{t} \left(-\frac{dt}{2}\right)$$

$$I = -4 \int_{1}^{\frac{1}{2}} \frac{1}{t} dt$$

We can eliminate the negative sign by reversing the upper and lower limits of integration.

$$I = 4 \int_{\frac{1}{2}}^{1} \frac{1}{t} dt$$

Integrate the expression.

$$I = 4 \left[ \ln\left|t\right| \right]_{\frac{1}{2}}^{1}$$

Substitute the upper and lower limits into the integrated expression.

$$I = 4 \left( \ln(1) - \ln\left(\frac{1}{2}\right) \right)$$

We know that $$\ln(1) = 0$$ and $$\ln\left(\frac{1}{2}\right) = -\ln(2)$$.

$$I = 4 (0 - (-\ln(2)))$$

$$I = 4 \ln(2)$$
$$I = \ln(16)$$

The final answer is $$\ln(16)$$.