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.

Well, using mathematica I can see that this function the given answer is the derivative of the integral. 
Consider 
\begin{align*}
f(x) &= \frac{\sqrt{2x^{4}-2x^{2}+1}}{x^{2}} \\ &= \frac{\frac{x^{2} \cdot (8x^{3}-4x)}{2 \sqrt{2x^{4}-2x^{2}+1}} - 2x \cdot \sqrt{2x^{4}-2x^{2}+1}}{x^{4}} \quad \ \Bigl[ \text{Note this is} \ f'(x)\Bigr] \\ &= \frac{x^{2} \cdot (4x^{3}-2x) - (2x^{4}-2x^{2}+1) \cdot 2x}{x^{4} \cdot \sqrt{2x^{4}-2x^{2}+1}} \\ &= \frac{4x^{4} -2x^{2} -4x^{4} + 4x^{2}-2}{x^{3} \cdot \sqrt{2x^{4}-2x^{2} +1}} \quad \Bigl[ \text{cancelling out x}\Bigr] \\ &= \frac{2 \cdot (x^{2}-1)}{x^{3} \cdot \sqrt{2x^{4}-2x^{2}+1}}
\end{align*}
So write your integral as $$\int\frac{x^{2}-1}{x^{3}\cdot \sqrt{2x^{4}-2x^{2}+1}} = \frac{1}{2} \int \frac{\text{d}}{\text{dx}}\biggl(\frac{\sqrt{2x^{4}-2x^{2}+1}}{x^{2}}\biggr)\ \text{dx}$$
Using this you can get the answer.
