Integral with $\sqrt{2x^4 - 2x^2 + 1}$ in the denominator $$\int\frac{x^{2}-1}{x^{3}\sqrt{2x^{4}-2x^{2}+1}} \: \text{d}x$$
I tried to substitute $x^2=t$ but I am unable to solve it and I also tried to divide numerator and denominator by $x^2$ and do something but could not get anything.

Substitute $u=1/x$ to get
$$
\int \frac{u^3 - u}{\sqrt{u^4-2u^2+2}}\,du
$$
This integral is much simpler, and can be solved by substituting $v = u^4 - 2u^2 + 2$.  The final result is
$$
\int\frac{x^{2}-1}{x^{3}\sqrt{2x^{4}-2x^{2}+1}}\,dx \;=\; \frac{\sqrt{2x^4-2x^2+1}}{2x^2} + C.
$$
