int frac{x}{sqrt{4 - x^4}} , dx = | vedclass

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(D) Let $I = \int \frac{x}{\sqrt{4 - x^4}} \, dx$. We can rewrite the integral as: $I = \int \frac{x}{\sqrt{2^2 - (x^2)^2}} \, dx$. Let $x^2 = t$. The

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$\frac{1}{2} \sin^{-1} \left( \frac{x^2}{2} \right)$

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(D) Let $I = \int \frac{x}{\sqrt{4 - x^4}} \, dx$. We can rewrite the integral as: $I = \int \frac{x}{\sqrt{2^2 - (x^2)^2}} \, dx$. Let $x^2 = t$. Then,differentiating both sides with respect to $x$,we get $2x \, dx = dt$,which implies $x \, dx = \frac{1}{2} \, dt$. Substituting these into the integral: $I = \int \frac{1}{\sqrt{2^2 - t^2}} \cdot \frac{1}{2} \, dt = \frac{1}{2} \int \frac{1}{\sqrt{2^2 - t^2}} \, dt$. Using the standard integral formula $\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \sin^{-1} \left( \frac{x}{a} \right) + C$,we get: $I = \frac{1}{2} \sin^{-1} \left( \frac{t}{2} \right) + C$. Substituting $t = x^2$ back,we get: $I = \frac{1}{2} \sin^{-1} \left( \frac{x^2}{2} \right) + C$.

$\int \frac{x}{\sqrt{4 - x^4}} \, dx = $

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