int frac{x^2-1}{x^3 sqrt{2 x^4-2 x^2+1}} d x= | vedclass

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(D) We have,$I = \int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} dx$. Divide the numerator and denominator inside the square root by $x^4$: $I = \int \fra

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$\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+c$

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(D) We have,$I = \int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} dx$. Divide the numerator and denominator inside the square root by $x^4$: $I = \int \frac{x^2-1}{x^3 \cdot x^2 \sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} dx = \int \frac{x^2-1}{x^5 \sqrt{2 - 2x^{-2} + x^{-4}}} dx$. $I = \int \frac{x^{-3} - x^{-5}}{\sqrt{2 - 2x^{-2} + x^{-4}}} dx$. Let $t = \sqrt{2 - 2x^{-2} + x^{-4}}$. Then $t^2 = 2 - 2x^{-2} + x^{-4}$. Differentiating both sides with respect to $x$: $2t \frac{dt}{dx} = 4x^{-3} - 4x^{-5} = 4(x^{-3} - x^{-5})$. So,$(x^{-3} - x^{-5}) dx = \frac{1}{2} t dt$. Substituting this into the integral: $I = \int \frac{\frac{1}{2} t dt}{t} = \frac{1}{2} \int dt = \frac{1}{2} t + c$. Substituting back $t$: $I = \frac{1}{2} \sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}} + c$.

$\int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x=$

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