If $\int_1^2 \frac{(x^2-1)dx}{x^3\sqrt{2x^4-2x^2+1}} = \frac{1}{k}$ then $k$ is____.
Step-by-Step Solution
Key Concept: Factor out $x^2$ from the radical to simplify the denominator and reveal the integrand structure.
To evaluate $I = \int_1^2 \frac{(x^2 - 1)dx}{x^2\sqrt{2 - 2x^{-2} + x^{-4}}}$, factor $x^2$ from the radical sign: $\sqrt{2 - 2x^{-2} + x^{-4}} = \frac{\sqrt{2x^4 - 2x^2 + 1}}{x^2}$. This simplifies the integral to $\int_1^2 \frac{(x^3 - x^{-1})dx}{\sqrt{2x^4 - 2x^2 + 1}}$. Recognizing the derivative structure and applying substitution methods yields the antiderivative in terms of the radical expression.
Correct Answer: 8