The value of int_0^1 tan^{-1}(1-x+x^2) dx is | vedclass

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(A) Let $I = \int_0^1 	an^{-1}(1-x+x^2) dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have: $I = \int_0^1 	an^{-1}(1-(1-x)+(1-x

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

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(A) Let $I = \int_0^1 	an^{-1}(1-x+x^2) dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have: $I = \int_0^1 	an^{-1}(1-(1-x)+(1-x)^2) dx$ $I = \int_0^1 	an^{-1}(1-1+x+1-2x+x^2) dx$ $I = \int_0^1 	an^{-1}(x^2-x+1) dx$. This shows the integral remains the same. We know that $	an^{-1}(1-x+x^2) = 	an^{-1}(1+x(x-1))$. Using $	an^{-1} A - 	an^{-1} B = 	an^{-1}(\frac{A-B}{1+AB})$,we can write: $1-x+x^2 = 1 + x(x-1)$. Thus,$	an^{-1}(1-x+x^2) = 	an^{-1}(1) + 	an^{-1}(x^2-x)$ is not directly applicable. Instead,note that $1-x+x^2 = 1 + x(x-1)$. Actually,$	an^{-1}(1-x+x^2) = 	an^{-1}(x) + 	an^{-1}(1-x)$. Integrating both sides: $I = \int_0^1 	an^{-1}(x) dx + \int_0^1 	an^{-1}(1-x) dx$. Since $\int_0^1 	an^{-1}(x) dx = \int_0^1 	an^{-1}(1-x) dx$,we have: $I = 2 \int_0^1 	an^{-1}(x) dx$. Using integration by parts: $\int 	an^{-1}(x) dx = x 	an^{-1}(x) - \frac{1}{2} \log(1+x^2)$. Evaluating from $0$ to $1$: $I = 2 [x 	an^{-1}(x) - \frac{1}{2} \log(1+x^2)]_0^1$ $I = 2 [ (1 \cdot \frac{\pi}{4} - \frac{1}{2} \log 2) - (0 - 0) ]$ $I = 2 [ \frac{\pi}{4} - \frac{1}{2} \log 2 ] = \frac{\pi}{2} - \log 2$.

The value of $\int_0^1 \tan^{-1}(1-x+x^2) dx$ is

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