The value of int frac{dx}{xsqrt{x^4 - 1}} is | vedclass

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

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$\frac{1}{2} \sec^{-1}(x^2) + k$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{x\sqrt{x^4 - 1}}$.Multiply the numerator and denominator by $x$:$I = \int \frac{x \, dx}{x^2 \sqrt{(x^2)^2 - 1}}$.Substitute $t = x^2$,then $dt = 2x \, dx$,which implies $x \, dx = \frac{dt}{2}$.Substituting these into the integral:$I = \int \frac{dt/2}{t \sqrt{t^2 - 1}} = \frac{1}{2} \int \frac{dt}{t \sqrt{t^2 - 1}}$.Using the standard integral formula $\int \frac{dt}{t \sqrt{t^2 - 1}} = \sec^{-1}(t) + k$:$I = \frac{1}{2} \sec^{-1}(t) + k$.Substituting back $t = x^2$:$I = \frac{1}{2} \sec^{-1}(x^2) + k$.

The value of $\int \frac{dx}{x\sqrt{x^4 - 1}}$ is

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