<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] 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 $\displaystyle 1000 \times \frac{u}{v} = 1000 \times \frac{1}{8} = 125$
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