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.

$$I = \int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}\,dx$$
Take $x^4$ factor out from the square root.
$$ = \int\frac{x^2-1}{x^5\sqrt{2-2x^{-2}+x^{-4}}}$$
Now substitute $\sqrt{2-2x^2+x^4}=t$
We have,
$$\,dt = \frac{4(x^{-3}-x^{-5})}{2\sqrt{2-2x^{-2}+x^{-4}}}\,dx$$
$$I = \frac{1}{2}\int dt$$
$$I = \frac{1}{2}\sqrt{2-2x^{-2}+x^{-4}}+C = \frac{\sqrt{2x^4-2x^2+1}}{2x^2}+C$$
