The value of int_0^1 tan ^{-1}left(frac{2 x-1 | vedclass

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(D) Let $I = \int_0^1 	an ^{-1}\left(\frac{2x-1}{1+x-x^2}ight) dx$.Using the property $	an^{-1}(a) - 	an^{-1}(b) = 	an^{-1}\left(\frac{a-b}{1+ab

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(D) Let $I = \int_0^1 	an ^{-1}\left(\frac{2x-1}{1+x-x^2}ight) dx$.Using the property $	an^{-1}(a) - 	an^{-1}(b) = 	an^{-1}\left(\frac{a-b}{1+ab}ight)$,we can rewrite the integrand:$\frac{2x-1}{1+x-x^2} = \frac{x - (1-x)}{1 + x(1-x)}$.Thus,$	an^{-1}\left(\frac{x-(1-x)}{1+x(1-x)}ight) = 	an^{-1}(x) - 	an^{-1}(1-x)$.So,$I = \int_0^1 	an^{-1}(x) dx - \int_0^1 	an^{-1}(1-x) dx$.Let $J = \int_0^1 	an^{-1}(1-x) dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get $J = \int_0^1 	an^{-1}(1-(1-x)) dx = \int_0^1 	an^{-1}(x) dx$.Therefore,$I = \int_0^1 	an^{-1}(x) dx - \int_0^1 	an^{-1}(x) dx = 0$.

The value of $\int_0^1 \tan ^{-1}\left(\frac{2 x-1}{1+x-x^2}\right) d x$ is

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