If 2 int_0^1 tan^{-1} x , dx = int_0^1 cot^{- | vedclass

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(B) We are given the identity $2 \int_{0}^{1} 	an^{-1} x \, dx = \int_{0}^{1} \cot^{-1} (1 - x + x^2) \, dx$. Using the property $\cot^{-1} u = \frac

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$\log 2$

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(B) We are given the identity $2 \int_{0}^{1} 	an^{-1} x \, dx = \int_{0}^{1} \cot^{-1} (1 - x + x^2) \, dx$. Using the property $\cot^{-1} u = \frac{\pi}{2} - 	an^{-1} u$,we can rewrite the right side: $2 \int_{0}^{1} 	an^{-1} x \, dx = \int_{0}^{1} \left( \frac{\pi}{2} - 	an^{-1} (1 - x + x^2) ight) dx$. $2 \int_{0}^{1} 	an^{-1} x \, dx = \frac{\pi}{2} - \int_{0}^{1} 	an^{-1} (1 - x + x^2) \, dx$. Rearranging the terms,we get: $\int_{0}^{1} 	an^{-1} (1 - x + x^2) \, dx = \frac{\pi}{2} - 2 \int_{0}^{1} 	an^{-1} x \, dx$. Now,we evaluate $I = \int_{0}^{1} 	an^{-1} x \, dx$ using integration by parts: $I = [x 	an^{-1} x]_0^1 - \int_0^1 \frac{x}{1+x^2} \, dx = \frac{\pi}{4} - \frac{1}{2} [\ln(1+x^2)]_0^1 = \frac{\pi}{4} - \frac{1}{2} \ln 2$. Substituting this back into our expression: $\int_{0}^{1} 	an^{-1} (1 - x + x^2) \, dx = \frac{\pi}{2} - 2 \left( \frac{\pi}{4} - \frac{1}{2} \ln 2 ight) = \frac{\pi}{2} - \frac{\pi}{2} + \ln 2 = \ln 2$.

If $2 \int_0^1 \tan^{-1} x \, dx = \int_0^1 \cot^{-1} (1 - x + x^2) \, dx,$ then $\int_0^1 \tan^{-1} (1 - x + x^2) \, dx$ is equal to

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