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Let $\mathrm{f}(\mathrm{x})$ be a function such that $\mathrm{f}^{\prime}\left(\frac{1}{\mathrm{x}}\right)+\mathrm{x}^{3} \mathrm{f}^{\prime}(\mathrm{x})=0$, What is
$\int_{-1}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}$ equal to? $\quad$
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$\int_{-1}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}$ equal to? $\quad$
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The correct answer is:
$2 \mathrm{f}(-1)$
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