If int frac{dx}{x (sqrt{x^4 - 1})} = frac{1}{ | vedclass

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(B) Let $I = \int \frac{dx}{x \sqrt{x^4 - 1}}$.Multiply the numerator and denominator by $x$:$I = \int \frac{x \ dx}{x^2 \sqrt{(x^2)^2 - 1}}$.Let $x^2

Vedclass · Accepted Answer

$2$

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(B) Let $I = \int \frac{dx}{x \sqrt{x^4 - 1}}$.Multiply the numerator and denominator by $x$:$I = \int \frac{x \ dx}{x^2 \sqrt{(x^2)^2 - 1}}$.Let $x^2 = t$,then $2x \ dx = dt$,or $x \ dx = \frac{dt}{2}$.Substituting these into the integral:$I = \int \frac{dt/2}{t \sqrt{t^2 - 1}} = \frac{1}{2} \int \frac{dt}{t \sqrt{t^2 - 1}}$.We know that $\int \frac{dt}{t \sqrt{t^2 - 1}} = \sec^{-1}(t) + C$.Therefore,$I = \frac{1}{2} \sec^{-1}(t) + C = \frac{1}{2} \sec^{-1}(x^2) + C$.Comparing this with the given expression $\frac{1}{k} \sec^{-1}(x^k)$,we get $k = 2$.

If $\int \frac{dx}{x (\sqrt{x^4 - 1})} = \frac{1}{k} \sec^{-1} (x^k)$,then the value of $k = $

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