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

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

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(A) Let $I = \int_{0}^{1} 	an ^{-1}\left[\frac{2 x-1}{1+x-x^{2}}ight] d x$. We can rewrite the argument of the $	an ^{-1}$ function as: $\frac{2x-1}{1+x-x^2} = \frac{x - (1-x)}{1 + x(1-x)}$. Using the property $	an^{-1} a - 	an^{-1} b = 	an^{-1} \left( \frac{a-b}{1+ab} ight)$,we get: $I = \int_{0}^{1} [	an^{-1} x - 	an^{-1}(1-x)] dx \quad \dots(1)$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we have: $I = \int_{0}^{1} [	an^{-1}(1-x) - 	an^{-1}(1-(1-x))] dx = \int_{0}^{1} [	an^{-1}(1-x) - 	an^{-1} x] dx \quad \dots(2)$. 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,$I = 0$.

$\int_{0}^{1} \tan ^{-1}\left[\frac{2 x-1}{1+x-x^{2}}\right] d x=$

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