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

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(B) 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

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(B) 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: $\frac{2x-1}{1+x-x^2} = \frac{x + (x-1)}{1 - x(x-1)}$. Using the identity $	an^{-1}(A) + 	an^{-1}(B) = 	an^{-1}\left(\frac{A+B}{1-AB}ight)$,we have: $I = \int_{0}^{1} (	an^{-1}(x) + 	an^{-1}(x-1)) d x$. Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,let $f(x) = 	an^{-1}(x-1)$. Then $\int_{0}^{1} 	an^{-1}(x-1) d x = \int_{0}^{1} 	an^{-1}((1-x)-1) d x = \int_{0}^{1} 	an^{-1}(-x) d x = -\int_{0}^{1} 	an^{-1}(x) d x$. Substituting this back into the integral: $I = \int_{0}^{1} 	an^{-1}(x) d x - \int_{0}^{1} 	an^{-1}(x) d x = 0$.

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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