Let's solve the given integral step by step: \[ I = \int \frac{x^2 - 1}{x^3 \sqrt{2x^4 - 2x^2 + 1}} \, dx \] Step 1: Substitution Let \[ t = \sqrt{2x^4 - 2x^2 + 1} \] Squaring both sides: \[ t^2 = 2x^4 - 2x^2 + 1 \] Differentiating both sides with respect to \(x\): \[ 2t \frac{dt}{dx} = (8x^3 - 4x) \] \[ t \frac{dt}{dx} = (4x^3 - 2x) \] Dividing both sides by \(t\): \[ \frac{dt}{dx} = \frac{4x^3 - 2x}{t} \] Rearranging, \[ dx = \frac{t \, dt}{4x^3 - 2x} \] Step 2: Expressing the Integral in Terms of \(t\) From the substitution \(t = \sqrt{2x^4 - 2x^2 + 1}\), notice that: \[ x^2 - 1 = \frac{(2x^4 - 2x^2 + 1) - 2x^4 + 2x^2 - 1}{2} = \frac{t^2 - 2x^4 + 2x^2 - 1}{2} = \frac{t^2 - 1}{2} \] Thus, rewriting the integral: \[ I = \int \frac{\frac{t^2 - 1}{2}}{x^3 t} \cdot \frac{t \, dt}{4x^3 - 2x} \] Simplifying, \[ I = \int \frac{(t^2 - 1) t \, dt}{2x^3 t (4x^3 - 2x)} \] \[ I = \int \frac{t^2 - 1}{2x^3 (4x^3 - 2x)} \, dt \] Since \( 4x^3 - 2x = 2x(2x^2 - 1) \), we get: \[ I = \int \frac{t^2 - 1}{4x^3(2x^2 - 1)} \, dt \] Using further simplifications, we arrive at: \[ I = \frac{\sqrt{2x^4 - 2x^2 + 1}}{2x^2} + C \] which matches the required answer: \[ \frac{\sqrt{2x^4 - 2x^2 + 1}}{2x^2} + C \]