TITLE: How to find $\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$
QUESTION [8 upvotes]: 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.

REPLY [4 votes]: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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With Euler Sub$\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}