int frac{1}{x sqrt{x^6+1}} , dx = | vedclass

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Question

(C) Let $I = \int \frac{1}{x \sqrt{x^6+1}} \, dx$. Multiply the numerator and denominator by $x^2$: $I = \int \frac{x^2}{x^3 \sqrt{(x^3)^2+1}} \, dx$.

Vedclass · Accepted Answer

$-\frac{1}{3} \operatorname{Sinh}^{-1}\left(\frac{1}{x^3}\right)+C$

Vedclass · Answer

(C) Let $I = \int \frac{1}{x \sqrt{x^6+1}} \, dx$. Multiply the numerator and denominator by $x^2$: $I = \int \frac{x^2}{x^3 \sqrt{(x^3)^2+1}} \, dx$. Let $u = x^3$,then $du = 3x^2 \, dx$,which implies $x^2 \, dx = \frac{du}{3}$. Substituting these into the integral: $I = \frac{1}{3} \int \frac{du}{u \sqrt{u^2+1}}$. Using the standard integral formula $\int \frac{du}{u \sqrt{u^2+a^2}} = -\frac{1}{a} \ln \left| \frac{a + \sqrt{u^2+a^2}}{u} \right| + C = -\frac{1}{a} \operatorname{Sinh}^{-1} \left( \frac{a}{u} \right) + C$: $I = \frac{1}{3} \left( -\operatorname{Sinh}^{-1} \left( \frac{1}{u} \right) \right) + C = -\frac{1}{3} \operatorname{Sinh}^{-1} \left( \frac{1}{x^3} \right) + C$.

$\int \frac{1}{x \sqrt{x^6+1}} \, dx =$

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