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If $2 \int \limits_0^1 \tan ^{-1} x dx = \int \limits _0^1 \cot ^{-1} (1-x+x^2)dx$ then $\int \limits_0^1 \tan ^{-1} (1-x+x^2)dx$ is equal to :

$\begin{array}{1 1} (1) \log 4 \\ (2) \large\frac{\pi}{2}+\log 2 \\ (3) \log 2 \\ (4) \large\frac{\pi}{2}- \log 4 \end{array} $

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