How to find $\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$ How to find ?$$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$$ 
I tried using the substitution $x^2=z$.But that did not help much.

A very long way... (Let $I$ equal the integral)
$$I=\frac{1}{\sqrt{2}}\int\frac{x^2-1}{x^3\sqrt{(x^2-1/2)^2+1/4}}dx$$
Let $(x^2-1/2)=u/2$ meaning $x=\sqrt{1/2(u+1)}$ and $dx=\frac{1}{4\sqrt{1/2(u+1)}}$ Hence
$$I=\frac{1}{\sqrt{2}}\int\frac{u-1}{(u+1)^2\sqrt{u^2+1}}du$$
Then let $u=\tan(v)$ and $du=\sec^2(v)$ giving
$$I=\frac{1}{\sqrt{2}}\int\frac{(\tan v -1)\sec v}{(\tan v+1)^2}dv$$
Using $\tan v=\frac{\sin v}{\cos v}$ and $\sec v=\frac{1}{\cos v}$ we have
$$I=\frac{1}{\sqrt{2}}\int\frac{\sin v-\cos v}{(\sin v +\cos v)^2}dv$$
Finally this can be evaluated using $m=\sin v + \cos v$ and $dm=\cos v - \sin v$ to get
$$I=-\frac{1}{\sqrt{2}}\int\frac{1}{m^2}dm=\frac{1}{\sqrt{2}m}$$
Doing substitutions back up to $x$ we have
$$I=\frac{1}{\sqrt{2}(\sin v + \cos v)}$$
$$=\frac{1}{\sqrt{2}(\sin(\arctan u)+\cos(\arctan u))}=\frac{\sqrt{u^2+1}}{\sqrt{2}\left(u + 1\right)}$$
$$=\frac{\sqrt{4 x^4-4 x^2+2}}{2
\sqrt{2}x^2}=\frac{\sqrt{2x^4-2x^2+1}}{2x^2}$$
