The integral of ${x^2 - 1 \over x^3 \sqrt{2x^4 - 2x^2 + 1}}$ 
$$\int {x^2 - 1 \over x^3 \sqrt{2x^4 - 2x^2 + 1}}dx$$


Substituting $u = x^2$, I get $${1\over 4}\int {u - 1 \over u^2 \sqrt{2u^2 - 2u + 1}}du$$ 
Then completing the square under the square root.
$${1\over 4}\int {u - 1 \over u^2 \sqrt{\left(\sqrt{2} u - \frac1{\sqrt{2}} \right)+ 1/2} }du$$ 
Substituting $\tan z = \sqrt{2}\left(\sqrt{2} u - \frac1{\sqrt{2}} \right)$
$${ \sqrt{2} \over 4}\int {\tan z -3 \over (\tan z- 1)^2 } \times \sec z \  dz$$
I think I can solve the last integral if I think a bit more but I need a better way to solve this question, this was seems very tedious.

Hint pull out $x^2$ common from numerator and $x^4$ in the root from denominator ie $x^2$ from numerator. Then put $1-\frac {1}{x^2}=u $ thus $\frac {2}{x^3}dx=du $ also write the new denominator as $1+(1-\frac {1}{x^2})^2$ hope you can handle  from  here.
