Let the integral be \( I = \int \frac{(x^2-1)dx}{x^3\sqrt{2x^4 - 2x^2 + 1}} \).
Divide the numerator and denominator by \( x^2 \). This seems complicated. Let's try dividing by a higher power. Let's manipulate the term inside the square root.
Factor out \( x^4 \) from the square root:
\[ \sqrt{2x^4 - 2x^2 + 1} = \sqrt{x^4(2 - \frac{2}{x^2} + \frac{1}{x^4})} = x^2\sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}} \]
Substitute this back into the integral:
\[ I = \int \frac{(x^2-1)dx}{x^3 \cdot x^2\sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} = \int \frac{(x^2-1)dx}{x^5\sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} \]
Now, let's split the numerator and divide by \( x^5 \):
\[ I = \int \frac{(\frac{x^2}{x^5} - \frac{1}{x^5})dx}{\sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} = \int \frac{(\frac{1}{x^3} - \frac{1}{x^5})dx}{\sqrt{2 - 2x^{-2} + x^{-4}}} \]
Let's use substitution. Let \( u = 2 - \frac{2}{x^2} + \frac{1}{x^4} = 2 - 2x^{-2} + x^{-4} \).
Now find the derivative, \( du \):
\[ \frac{du}{dx} = -2(-2)x^{-3} + (-4)x^{-5} = 4x^{-3} - 4x^{-5} = 4(\frac{1}{x^3} - \frac{1}{x^5}) \]
So, \( (\frac{1}{x^3} - \frac{1}{x^5})dx = \frac{du}{4} \).
Substitute \( u \) and \( du \) back into the integral:
\[ I = \int \frac{1}{\sqrt{u}} \frac{du}{4} = \frac{1}{4} \int u^{-1/2} du \]
\[ I = \frac{1}{4} \frac{u^{1/2}}{1/2} + C = \frac{1}{2} u^{1/2} + C = \frac{1}{2}\sqrt{u} + C \]
Substitute back the expression for \( u \):
\[ I = \frac{1}{2}\sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}} + C \]
This matches option (C).