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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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$\begin{aligned} I &=\int_{0}^{1} \tan ^{-1}\left[\frac{2 x-1}{1+x-x^{2}}\right] d x \\ &=\int_{0}^{1} \tan ^{-1}\left[\frac{2 x-1}{1+x(1-x)}\right] d x=\int_{0}^{1} \tan ^{-1}\left[\frac{x-(1-x)}{1+x(1-x)}\right] d x \\ &=\int_{0}^{1}\left[\tan ^{-1} x-\tan ^{-1}(1-x)\right] d x ...(1) \\ &=\int_{0}^{1}\left\{\tan ^{-1}(1-x)-\tan ^{-1}[1-(1-x)]\right\} d x \\ &=\int_{0}^{1}\left[\tan ^{-1}(1-x)-\tan ^{-1} x\right] d x ...(2) \\ \text { Equation }(1)+(2) \operatorname{gives} \\ 2 I=0 \Rightarrow I=0 \end{aligned}$
$\begin{aligned} I &=\int_{0}^{1} \tan ^{-1}\left[\frac{2 x-1}{1+x-x^{2}}\right] d x \\ &=\int_{0}^{1} \tan ^{-1}\left[\frac{2 x-1}{1+x(1-x)}\right] d x=\int_{0}^{1} \tan ^{-1}\left[\frac{x-(1-x)}{1+x(1-x)}\right] d x \\ &=\int_{0}^{1}\left[\tan ^{-1} x-\tan ^{-1}(1-x)\right] d x ...(1) \\ &=\int_{0}^{1}\left\{\tan ^{-1}(1-x)-\tan ^{-1}[1-(1-x)]\right\} d x \\ &=\int_{0}^{1}\left[\tan ^{-1}(1-x)-\tan ^{-1} x\right] d x ...(2) \\ \text { Equation }(1)+(2) \operatorname{gives} \\ 2 I=0 \Rightarrow I=0 \end{aligned}$
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