If int_{0}^{1} tan ^{-1} x , dx = p,then the | vedclass

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(B) Let $I = \int_{0}^{1} 	an ^{-1}\left(\frac{1-x}{1+x}ight) \, dx$. Using the trigonometric identity $	an ^{-1} A - 	an ^{-1} B = 	an ^{-1}\le

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$\frac{\pi}{4} - p$

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(B) Let $I = \int_{0}^{1} 	an ^{-1}\left(\frac{1-x}{1+x}ight) \, dx$. Using the trigonometric identity $	an ^{-1} A - 	an ^{-1} B = 	an ^{-1}\left(\frac{A-B}{1+AB}ight)$,we can write $	an ^{-1}\left(\frac{1-x}{1+x}ight) = 	an ^{-1}(1) - 	an ^{-1}(x)$. Since $	an ^{-1}(1) = \frac{\pi}{4}$,the integral becomes: $I = \int_{0}^{1} \left( \frac{\pi}{4} - 	an ^{-1}(x) ight) \, dx$. $I = \int_{0}^{1} \frac{\pi}{4} \, dx - \int_{0}^{1} 	an ^{-1}(x) \, dx$. $I = \left[ \frac{\pi}{4} x ight]_{0}^{1} - p$. $I = \frac{\pi}{4}(1 - 0) - p$. $I = \frac{\pi}{4} - p$.

If $\int_{0}^{1} \tan ^{-1} x \, dx = p$,then the value of $\int_{0}^{1} \tan ^{-1}\left(\frac{1-x}{1+x}\right) \, dx$ is

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