int frac{x}{sqrt{1-2 x^4}} , dx = (Where C i | vedclass

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(A) Let $I = \int \frac{x}{\sqrt{1-2 x^4}} \, dx$. Substitute $u = \sqrt{2} x^2$. Then $du = 2 \sqrt{2} x \, dx$,which implies $x \, dx = \frac{du}{2

Vedclass · Accepted Answer

$\frac{1}{2 \sqrt{2}} \sin^{-1}(\sqrt{2} x^2) + C$

Vedclass · Answer

(A) Let $I = \int \frac{x}{\sqrt{1-2 x^4}} \, dx$. Substitute $u = \sqrt{2} x^2$. Then $du = 2 \sqrt{2} x \, dx$,which implies $x \, dx = \frac{du}{2 \sqrt{2}}$. Substituting these into the integral: $I = \int \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{2 \sqrt{2}} = \frac{1}{2 \sqrt{2}} \int \frac{du}{\sqrt{1-u^2}}$. Using the standard integral formula $\int \frac{1}{\sqrt{1-u^2}} \, du = \sin^{-1}(u) + C$,we get: $I = \frac{1}{2 \sqrt{2}} \sin^{-1}(u) + C = \frac{1}{2 \sqrt{2}} \sin^{-1}(\sqrt{2} x^2) + C$.

$\int \frac{x}{\sqrt{1-2 x^4}} \, dx = $ (Where $C$ is a constant of integration)

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