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$\int \frac{x}{\sqrt{1-2 x^4}} \mathrm{~d} x=$ (Where $C$ is a constant of integration)
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The correct answer is:
$\frac{1}{2 \sqrt{2}} \sin ^{-1}\left(\sqrt{2} x^2\right)+C$
$\int \frac{x}{\sqrt{1-2 x^4}} \mathrm{~d} x=\frac{1}{2 \sqrt{2}} \int \frac{2 \sqrt{2} x \mathrm{~d} x}{\sqrt{1-\left(\sqrt{2} x^2\right)^2}}=\frac{1}{2 \sqrt{2}} \sin ^{-1}\left(\sqrt{2} x^2\right)$
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