A) 1
B) 0
C) \[-1\]
D) None of these
Correct Answer: B
Solution :
\[I=\int_{0}^{1}{{{\tan }^{-1}}}\left( \frac{2x-1}{1+x-{{x}^{2}}} \right)\,dx\]\[=\int_{0}^{1}{{{\tan }^{-1}}}\left( \frac{x+(x-1)}{1-x(x-1)} \right)\,dx\] \[I=\int_{0}^{1}{({{\tan }^{-1}}x+{{\tan }^{-1}}(x-1))}\,dx\] \[I=\int_{0}^{1}{{{\tan }^{-1}}x\,dx+\int_{0}^{1}{{{\tan }^{-1}}(x-1)\,dx}}\] \[I=\int_{0}^{1}{{{\tan }^{-1}}x\,dx+\int_{0}^{1}{{{\tan }^{-1}}(1-x-1)\,dx}}\], {Using \[\int_{0}^{a}{f(x)dx=\int_{0}^{a}{f(a-x)dx}}\]in second integral} \[I=\int_{0}^{1}{{{\tan }^{-1}}x\,dx+\int_{0}^{1}{{{\tan }^{-1}}(-x)\,dx}}\] \[I=\int_{0}^{1}{{{\tan }^{-1}}x\,dx-\int_{0}^{1}{{{\tan }^{-1}}x\,dx}=0}\].
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