Find $\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx$ $$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\frac{u}{v}$$ where $u$ and $v$ are in their lowest form. Find the value of $\dfrac{1000u}{v}$
$$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^2(x^2-1)+1}}dx$$ I put $x^2-1=t$ but no benefit. Please guide me.
 A: Given $$\displaystyle  I = \int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx = \int_{1}^{2}\frac{x^2-1}{x^3\cdot x^2\sqrt{2-2x^{-2}+x^{-4}}}dx$$
$$\displaystyle = \int_{1}^{2}\frac{x^{-3}-x^{-5}}{\sqrt{2-2x^{-2}+x^{-4}}}dx$$
Now Let $2-2x^{-2}+x^{-4} = t^2\;,$ Then $\displaystyle \left(x^{-3}-x^{-5}\right)dx = \frac{2t}{4}dt$
so we Get $$\displaystyle I = \frac{1}{2}\int_{1}^{2}\frac{t}{t}dt = \frac{1}{2}t = \frac{1}{2}\left[\sqrt{2-2x^{-2}+x^{-4}}\right]_{1}^{2} = \frac{1}{2}\left[\frac{\sqrt{2x^4-2x^2+1}}{x^2}\right]_{1}^{2}$$
So we get $$\displaystyle I = \int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx = \frac{1}{8}$$
But Here Given $$\displaystyle I = \frac{1}{8} = \frac{u}{v}.$$ So we get $\displaystyle u=1\;, v=8$
So  we get $$\displaystyle \frac{1000u}{v} = \frac{1000 \times 1}{8} = 125.$$
A: Notice, we have $$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt {2x^4-2x^2+1}}dx$$
 $$=\int_{1}^{2}\frac{x^2-1}{x^3\sqrt {2\left(x^2-\frac{1}{2}\right)^2-\frac{1}{2}+1}}dx$$
$$=\frac{1}{2}\int_{1}^{2}\frac{2x(x^2-1)}{x^4\sqrt {2\left(x^2-\frac{1}{2}\right)^2+\frac{1}{2}}}dx$$ Now, $$x^2-\frac{1}{2}=t\implies 2xdx=dt$$
$$=\frac{1}{2}\int_{1/2}^{7/2}\frac{\left(t-\frac{1}{2}\right)dt}{\left(t-\frac{1}{2}\right)^2\sqrt {2t^2+\frac{1}{2}}}dx$$
$$=\frac{1}{2}\int_{1/2}^{7/2}\frac{dt}{\left(t-\frac{1}{2}\right)\sqrt {2t^2+\frac{1}{2}}}dx$$
I hope you can take it from here.
A: A couple of hints (not a full solution):
$x\to 2 \sin x$
\begin{align}
\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}\,dx&=\int_{\pi/6}^{\pi/2}\frac{\cot (x) \left(3-\cot ^2(x)\right)}{4 \sqrt{-12 \cos (2 x)+4 \cos (4 x)+9}}\,dx\\
\end{align}
now you may use  $$\cot (x) \left(3-\cot ^2(x)\right)=\Big[\frac14(24 \sin (2 x)-16 \sin (4 x)) \csc ^2(x)\Big]-\Big[(-12 \cos (2 x)+4 \cos (4 x)+9) \cot (x) \csc ^2(x)\Big]$$
A: Let $$\displaystyle I = \int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx = \int \frac{\left(1-x^{-2}\right)\cdot x^{-3}}{\sqrt{2-2x^{-2}+x^{-2}}}dx\;,$$ 
Now Let $x^{-2} = u\;,$ Then $-2x^{-3}dx = du$ and Changing Limit, We get
$$\displaystyle I = -\frac{1}{2}\int_{1}^{\frac{1}{4}}\frac{1-u}{\sqrt{u^2-2u+2}}du = -\frac{1}{2}\int_{1}^{\frac{1}{4}}\frac{(1-u)}{\sqrt{(1-u)^2+1}}du$$
Now Let $(1-u) = t\;,$ Then $du = -dt$ and Changing Limit, We get 
$$\displaystyle I = \frac{1}{2}\int_{0}^{\frac{3}{4}}\frac{t}{\sqrt{1+t^2}}dt=\frac{1}{2}\left[\sqrt{1+t^2}\right]_{0}^{\frac{3}{4}} =\frac{1}{8}$$
So we get $\displaystyle I = \frac{1}{8} = \frac{u}{v},$ So $u=1,v=8$
So we get $\displaystyle 1000 \times \frac{u}{v} = 1000 \times \frac{1}{8} = 125$
