int_{0}^{1} tan ^{-1}left(frac{2x-1}{1+x-x^{2 | vedclass

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(A) Let $I = \int_{0}^{1} 	an ^{-1}\left(\frac{2x-1}{1+x-x^{2}}ight) dx$. We can rewrite the argument of the inverse tangent function as: $\frac{2x

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(A) Let $I = \int_{0}^{1} 	an ^{-1}\left(\frac{2x-1}{1+x-x^{2}}ight) dx$. We can rewrite the argument of the inverse tangent function as: $\frac{2x-1}{1+x-x^{2}} = \frac{x - (1-x)}{1 + x(1-x)}$. Using the identity $	an^{-1}(a) - 	an^{-1}(b) = 	an^{-1}\left(\frac{a-b}{1+ab}ight)$,we have: $I = \int_{0}^{1} (	an^{-1} x - 	an^{-1}(1-x)) dx$ ... $(1)$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get: $I = \int_{0}^{1} (	an^{-1}(1-x) - 	an^{-1}(1-(1-x))) dx = \int_{0}^{1} (	an^{-1}(1-x) - 	an^{-1} x) dx$ ... $(2)$. Adding equations $(1)$ and $(2)$: $2I = \int_{0}^{1} (	an^{-1} x - 	an^{-1}(1-x) + 	an^{-1}(1-x) - 	an^{-1} x) dx = \int_{0}^{1} 0 dx = 0$. Therefore,$I = 0$.

$\int_{0}^{1} \tan ^{-1}\left(\frac{2x-1}{1+x-x^{2}}\right) dx =$

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