<s>[INST] 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.
 [/INST] 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 $$\displaystyle \frac{1000u}{v} = \frac{1000 \times 1}{8} = 125.$$
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