Integrate $\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$ Evaluate $$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$$
I tried $u=\sqrt{2x^4-2x^2+1}$, $u=\dfrac{1}{x}$ and $u=\sqrt{x}$ but none of these worked.
A friend gave this and said it's from IIT JEE and answer is $$\dfrac{\sqrt{2x^4-2x^2+1}}{2x^2}+C$$
I would like a hint or suggestion.

Let $I=\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$. Divide Numerator and denominator of integrand by $x^2$ to get: $I=\int \frac{x^{-3}-x^{-5}}{\sqrt{2-2x^{-2}+x^{-4}}} \mathop{dx}$. Substitute $y=2-2x^{-2}+x^{-4}$ so that $dy=4(x^{-3}-x^{-5})dx$ and hence $I=\int \frac{1}{4 \sqrt{y}} \mathop{dy}=\frac{y^{1/2}}{2}+C=\frac{\sqrt{2-2x^{-2}+x^{-4}}}{2}+C$, where $C$ is integration constant.
