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

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(B) We have the integral $I = \int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x$.Divide the numerator and denominator by $x^5$ inside the square root:$I

Vedclass · Accepted Answer

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

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(B) We have the integral $I = \int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x$.Divide the numerator and denominator by $x^5$ inside the square root:$I = \int \frac{\frac{x^2-1}{x^5}}{\sqrt{\frac{2 x^4-2 x^2+1}{x^8}}} d x = \int \frac{\frac{1}{x^3}-\frac{1}{x^5}}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}} d x$.Let $t = 2-\frac{2}{x^2}+\frac{1}{x^4}$. Then $dt = (\frac{4}{x^3}-\frac{4}{x^5}) dx$,which implies $(\frac{1}{x^3}-\frac{1}{x^5}) dx = \frac{dt}{4}$.Substituting these into the integral:$I = \int \frac{dt}{4 \sqrt{t}} = \frac{1}{4} \int t^{-1/2} dt = \frac{1}{4} \cdot \frac{t^{1/2}}{1/2} + C = \frac{1}{2} \sqrt{t} + C$.Substituting $t$ back,we get $I = \frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}} + C = \frac{1}{2} \sqrt{\frac{2x^4-2x^2+1}{x^4}} + C = \frac{1}{2x^2} \sqrt{2x^4-2x^2+1} + C$.

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

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