The value of int_0^1 {{{tan }^{ - 1}}left( {f | vedclass

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(B) 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}$ functio

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(B) 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:$\frac{2x - 1}{1 + x - x^2} = \frac{x + (x - 1)}{1 - x(x - 1)}$.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}(x - 1)) \,dx$.Now,split the integral:$I = \int_0^1 	an^{-1}x \,dx + \int_0^1 	an^{-1}(x - 1) \,dx$.Apply the property $\int_0^a f(x) \,dx = \int_0^a f(a - x) \,dx$ to the second integral:Let $J = \int_0^1 	an^{-1}(x - 1) \,dx$. Substituting $x = 1 - t$,$dx = -dt$:$J = \int_1^0 	an^{-1}(1 - t - 1) \,(-dt) = \int_0^1 	an^{-1}(-t) \,dt = -\int_0^1 	an^{-1}t \,dt$.Thus,$I = \int_0^1 	an^{-1}x \,dx - \int_0^1 	an^{-1}x \,dx = 0$.

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

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