int tan ^{-1}left(1-x+x^2right) d x+int tan ^ | vedclass

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(A) Let $I = \int 	an ^{-1}(1-x+x^2) dx + \int 	an ^{-1}(x) dx + \int 	an ^{-1}(1-x) dx$. Using the property $	an^{-1}(A) + 	an^{-1}(B) = 	an^{-

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$\frac{\pi}{2} x+C$

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(A) Let $I = \int 	an ^{-1}(1-x+x^2) dx + \int 	an ^{-1}(x) dx + \int 	an ^{-1}(1-x) dx$. Using the property $	an^{-1}(A) + 	an^{-1}(B) = 	an^{-1}\left(\frac{A+B}{1-AB}ight)$,we combine the last two terms: $	an^{-1}(x) + 	an^{-1}(1-x) = 	an^{-1}\left(\frac{x + 1 - x}{1 - x(1-x)}ight) = 	an^{-1}\left(\frac{1}{1-x+x^2}ight)$. Since $	an^{-1}\left(\frac{1}{u}ight) = \cot^{-1}(u)$ for $u > 0$,we have: $I = \int \left[ 	an^{-1}(1-x+x^2) + \cot^{-1}(1-x+x^2) ight] dx$. Using the identity $	an^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$: $I = \int \frac{\pi}{2} dx = \frac{\pi}{2} x + C$.

$\int \tan ^{-1}\left(1-x+x^2\right) d x+\int \tan ^{-1}(x) d x+\int \tan ^{-1}(1-x) d x=$

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