The value of int frac{(x^2-1) dx}{x^3 sqrt{2x | vedclass

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

Vedclass · Accepted Answer

$\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+c$,where $c$ is a constant of integration.

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(C) Let $I = \int \frac{(x^2-1) dx}{x^3 \sqrt{2x^4-2x^2+1}}$.Divide the numerator and denominator inside the square root by $x^4$:$I = \int \frac{(x^2-1) dx}{x^3 \cdot x^2 \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}} = \int \frac{(\frac{1}{x^3} - \frac{1}{x^5}) dx}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}$.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 $I = \frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}} + c$.

The value of $\int \frac{(x^2-1) dx}{x^3 \sqrt{2x^4-2x^2+1}}$ is

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