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.

Solution 1:

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 $$ \int\frac{x^{2}-1}{x^{3}\sqrt{2x^{4}-2x^{2}+1}}\,dx \;=\; \frac{\sqrt{2x^4-2x^2+1}}{2x^2} + C. $$

Solution 2:

The factor of $(2x^4 - 2x^2 + 1)^{-1/2}$ suggests that it might be profitable to look at solutions of the form $f(x)(2x^4 - 2x^2 + 1)^{1/2}$, and hope for a simplification. Indeed, by differentiating this expression we get the differential equation

$$(2x^4 - 2x^2 + 1) \frac{\mathrm{d}f}{\mathrm{d}x} + 2x(2x^2 - 1) f = \frac{x^2-1}{x^3}$$

which can be solved with an integrating factor.