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$\int_{-\pi}^{\pi} \frac{2 x}{1+\cos ^{2} x} d x=$
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Verified Answer
The correct answer is:
$0$
Let $f(x)=\frac{2 x}{1+\cos ^{2} x} \Rightarrow f(-x)=\frac{-2 x}{1+\cos ^{2} x}$
Thus $\mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{I}=0$
Thus $\mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{I}=0$
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