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$\int_0^1 \frac{d}{d x}\left[\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right] d x$ is equal to
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$\pi / 2$
$I=\left[\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right]_0^1=\sin ^{-1}(1)-\sin ^{-1}(0)=\frac{\pi}{2}$
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