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

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(D) Let $I = \int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x$. Divide the numerator and the denominator by $x^5$: $I = \int \frac{\frac{1}{x^3} - \fra

Vedclass · Accepted Answer

$\frac{\left(1-2 x^2+2 x^4\right)^{\frac{1}{2}}}{2 x^2}+c$

Vedclass · Answer

(D) Let $I = \int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x$. Divide the numerator and the denominator by $x^5$: $I = \int \frac{\frac{1}{x^3} - \frac{1}{x^5}}{\sqrt{2 - \frac{2}{x^2} + \frac{1}{x^4}}} d x$. Let $t = 2 - \frac{2}{x^2} + \frac{1}{x^4}$. Then $dt = (\frac{4}{x^3} - \frac{4}{x^5}) d x$,which implies $(\frac{1}{x^3} - \frac{1}{x^5}) d x = \frac{dt}{4}$. Substituting these into the integral: $I = \frac{1}{4} \int \frac{1}{\sqrt{t}} dt = \frac{1}{4} (2\sqrt{t}) + c = \frac{\sqrt{t}}{2} + c$. Substituting $t = 2 - \frac{2}{x^2} + \frac{1}{x^4} = \frac{2x^4 - 2x^2 + 1}{x^4}$ back: $I = \frac{1}{2} \sqrt{\frac{2x^4 - 2x^2 + 1}{x^4}} + c = \frac{\sqrt{2x^4 - 2x^2 + 1}}{2x^2} + c$.

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

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