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$\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x=$
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$\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x$
$=\int_{0}^{1} \tan ^{-1}\left[\frac{x-(1-x)}{1+x(1-x)}\right] d x=\int_{0}^{1}\left[\tan ^{-1} x-\tan ^{-1}(1-x)\right] d x$...(1)
$=\int_{0}^{1}\left[\tan ^{-1}(1-x)-\tan ^{-1}(1-(1-x))\right] d x=\int_{0}^{1}\left[\tan ^{-1}(1-x)-\tan ^{-1}(x)\right] d x$...(2)
Adding (1) \& (2), we get
$2 I=0 \Rightarrow I=0$
$\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x$
$=\int_{0}^{1} \tan ^{-1}\left[\frac{x-(1-x)}{1+x(1-x)}\right] d x=\int_{0}^{1}\left[\tan ^{-1} x-\tan ^{-1}(1-x)\right] d x$...(1)
$=\int_{0}^{1}\left[\tan ^{-1}(1-x)-\tan ^{-1}(1-(1-x))\right] d x=\int_{0}^{1}\left[\tan ^{-1}(1-x)-\tan ^{-1}(x)\right] d x$...(2)
Adding (1) \& (2), we get
$2 I=0 \Rightarrow I=0$
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