The value of int_{0}^{1} tan ^{-1}left(frac{2 | vedclass

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

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$0$

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(D) Let $I = \int_{0}^{1} 	an^{-1}\left(\frac{2x-1}{1+x-x^2}ight) dx$.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 identity $	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$ ...... $(1)$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$$I = \int_{0}^{1} [	an^{-1}(1-x) - 	an^{-1}(x)] dx$ ...... $(2)$Adding $(1)$ and $(2)$:$2I = \int_{0}^{1} [	an^{-1}(x) - 	an^{-1}(1-x) + 	an^{-1}(1-x) - 	an^{-1}(x)] dx$$2I = \int_{0}^{1} 0 dx = 0$.Therefore,$I = 0$.Hence,the correct answer is $D$.

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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