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: By setting $x^2=z$ we are left with:
$$ \frac{1}{2}\int\frac{z-1}{z^2\sqrt{2z^2-2z+1}}\,dz=C+\frac{\sqrt{2z^2-2z+1}}{2z}.$$
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With Euler Sub$\ldots$
$$
x = {\root{\root{2} - \root{2}t^{2}} \over \root{2}\root{\root{2} + 2t}}
\quad\imp\quad
t \equiv \root{2x^{4} - 2x^{2} + 1} - \root{2}x^{2}
$$
the integral adopt a relatively simple form
\begin{align}
\int{x^{2} - 1 \over x^{3}\root{2x^{4} - 2x^{2} + 1}}\,\dd x & =
\int{t^{2} + 2\root{2}t + 1 \over \pars{t^{2} - 1}^{2}}\,\dd t
\\[3mm] & =
\int{\dd t \over t^{2} - 1}\,\dd t +
\root{2}\int{2t\,\dd t \over t^{2} - 1}\,\dd t +
2\int{\dd t  \over \pars{t^{2} - 1}^{2}} = -\,{t + \root{2} \over t^{2} - 1} 
\end{align}
A: 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
$$I=\frac{1}{\sqrt{2}(\sin v + \cos v)}$$
$$=\frac{1}{\sqrt{2}(\sin(\arctan u)+\cos(\arctan u))}=\frac{\sqrt{u^2+1}}{\sqrt{2}\left(u + 1\right)}$$
$$=\frac{\sqrt{4 x^4-4 x^2+2}}{2
\sqrt{2}x^2}=\frac{\sqrt{2x^4-2x^2+1}}{2x^2}$$
A: Let $$I = \int\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx = \int\frac{x^2-1}{x^3\cdot x^2\sqrt{2-2x^{-2}+x^{-4}}}dx$$
So $$I=\int\frac{x^{-3}-x^{-5}}{\sqrt{2-2x^{-2}+x^{-4}}}dx$$
Now Put $2-2x^{-2}+x^{-4} = t^2\;,$ Then $4(x^{-3}-x^{-5})dx = 2tdt$
So $$I = \frac{1}{2}\int\frac{t}{t}dt = \frac{1}{2}t+\mathcal{C}=\frac{1}{2}\sqrt{2-2x^{-2}+x^{-4}}+\mathcal{C}$$
So $$I = \frac{\sqrt{2x^4-2x^2+1}}{2x^2}+\mathcal{C}$$
A: You can apply $u=x^2$ :
$$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx=\frac{1}{2}\int \frac{x^2-1}{x^4\sqrt{2x^4-2x^2+1}} 2xdx=\frac{1}{2}\int \frac{u-1}{u^2\sqrt{2u^2-2u+1}} du$$
Now note that :
$$\int \frac{u-1}{u^2\sqrt{2u^2-2u+1}} du=\int \frac{u-1}{u^2\sqrt{2(u^2-u)+1}} du$$
we can substitute $v=\frac{1}{u}$ :
$$\int \frac{u-1}{u^2\sqrt{2(u^2-u)+1}} du=\int \frac{1-u}{\sqrt{2(u^2-u)+1}} \frac{-du}{u^2}=\int \frac{1-\frac{1}{v}}{\sqrt{2(\frac{1}{v^2}-\frac{1}{v})+1}} dv\\=\int \frac{\frac{v-1}{v}}{\sqrt{\frac{2-2v+v^2}{v^2}}}dv=\int \frac{v-1}{\sqrt{v^2-2v+2}}dv$$
Substitute $y=v^2-2v+2$ and you should be able to finish.
