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

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(D) Let $I = \int_{0}^{1} 	an^{-1} \left( \frac{1}{x^2 - x + 1} ight) dx$. Using the identity $	an^{-1} a - 	an^{-1} b = 	an^{-1} \left( \frac{a

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$\frac{\pi}{2} - \ln 2$

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(D) Let $I = \int_{0}^{1} 	an^{-1} \left( \frac{1}{x^2 - x + 1} ight) dx$. Using the identity $	an^{-1} a - 	an^{-1} b = 	an^{-1} \left( \frac{a - b}{1 + ab} ight)$,we can write: $\frac{1}{x^2 - x + 1} = \frac{x - (x - 1)}{1 + x(x - 1)}$. Thus,$	an^{-1} \left( \frac{1}{x^2 - x + 1} ight) = 	an^{-1} x - 	an^{-1} (x - 1)$. $I = \int_{0}^{1} 	an^{-1} x dx - \int_{0}^{1} 	an^{-1} (x - 1) dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$ for the second integral: $\int_{0}^{1} 	an^{-1} (x - 1) dx = \int_{0}^{1} 	an^{-1} ((1 - x) - 1) dx = \int_{0}^{1} 	an^{-1} (-x) dx = -\int_{0}^{1} 	an^{-1} x dx$. Substituting this back into $I$: $I = \int_{0}^{1} 	an^{-1} x dx - (-\int_{0}^{1} 	an^{-1} x dx) = 2 \int_{0}^{1} 	an^{-1} x dx$. Using integration by parts: $\int 	an^{-1} x dx = x 	an^{-1} x - \frac{1}{2} \ln(1 + x^2)$. $I = 2 [x 	an^{-1} x - \frac{1}{2} \ln(1 + x^2)]_{0}^{1} = 2 [ (1 \cdot \frac{\pi}{4} - \frac{1}{2} \ln 2) - (0 - 0) ] = 2 [ \frac{\pi}{4} - \frac{1}{2} \ln 2 ] = \frac{\pi}{2} - \ln 2$.

The value of $\int_{0}^{1} \tan^{-1} \left( \frac{1}{x^2 - x + 1} \right) dx$ is

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