Question:

Step 1: Rewrite the integrand to prepare for a suitable substitution. The given integral is: $$ I = \int \frac{x^2 - 1}{x^3 \sqrt{2x^4 - 2x^2 + 1}} dx $$ To simplify the expression inside the square root, factor out $x^4$: $$ \sqrt{2x^4 - 2x^2 + 1} = \sqrt{x^4 \left(2 - \frac{2}{x^2} + \frac{1}{x^4}\right)} = x^2 \sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}} $$ Substitute this back into the integral: $$ I = \int \frac{x^2 - 1}{x^3 \cdot x^2 \sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} dx = \int \frac{x^2 - 1}{x^5 \sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} dx $$ Now, divide the numerator by $x^5$ to express it in terms of powers of $1/x$: $$ I = \int \frac{\frac{x^2 - 1}{x^5}}{\sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} dx = \int \frac{\frac{1}{x^3} - \frac{1}{x^5}}{\sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} dx $$ Step 2: Introduce a substitution. Let $t$ be the expression inside the square root: $$ t = 2 - \frac{2}{x^2} + \frac{1}{x^4} $$ Step 3: Calculate the differential $dt$. Differentiate $t$ with respect to $x$: $$ \frac{dt}{dx} = \frac{d}{dx} \left(2 - 2x^{-2} + x^{-4}\right) $$ $$ \frac{dt}{dx} = 0 - 2(-2)x^{-3} + (-4)x^{-5} = \frac{4}{x^3} - \frac{4}{x^5} $$ So, the differential $dt$ is: $$ dt = \left(\frac{4}{x^3} - \frac{4}{x^5}\right) dx = 4 \left(\frac{1}{x^3} - \frac{1}{x^5}\right) dx $$ Rearrange to find the expression in the numerator of our integral: $$ \left(\frac{1}{x^3} - \frac{1}{x^5}\right) dx = \frac{1}{4} dt $$ Step 4: Substitute $t$ and $dt$ into the integral and evaluate. Substitute $t$ and the expression for $dt$ into the integral $I$: $$ I = \int \frac{1}{\sqrt{t}} \cdot \frac{1}{4} dt = \frac{1}{4} \int t^{-1/2} dt $$ Now, integrate with respect to $t$: $$ I = \frac{1}{4} \left(\frac{t^{-1/2 + 1}}{-1/2 + 1}\right) + c = \frac{1}{4} \left(\frac{t^{1/2}}{1/2}\right) + c $$ $$ I = \frac{1}{4} \cdot 2\sqrt{t} + c = \frac{1}{2}\sqrt{t} + c $$ Step 5: Substitute back for $t$ to obtain the final answer and match with the options. Substitute $t = 2 - \frac{2}{x^2} + \frac{1}{x^4}$ back into the expression for $I$: $$ I = \frac{1}{2}\sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}} + c $$ To match the given options, combine the terms inside the square root and simplify: $$ I = \frac{1}{2}\sqrt{\frac{2x^4 - 2x^2 + 1}{x^4}} + c = \frac{1}{2} \frac{\sqrt{2x^4 - 2x^2 + 1}}{\sqrt{x^4}} + c $$ Since $\sqrt{x^4} = x^2$ (for $x \neq 0$), the integral becomes: $$ I = \frac{\sqrt{2x^4 - 2x^2 + 1}}{2x^2} + c $$ Comparing this with the given options, the result matches Option D. The final answer is $\boxed{\text{(D)}}$.