∫01Tan−12x−11+x−x2dx=

I=∫01tan−1(2x−11+x−x2)dx=∫01tan−1(x−(1−x)1+x(1−x))dx=∫01[tan−1x−tan−1(1−x)]dx[∴ tan−1(x−y1+xy)=tan−1x−tan−1y]=∫01tan−1x−∫01tan−1(1−1-x)dx=0[∵∫0af(x)dx=∫0af(a−x)dx]

∫01Tan−12x−11+x−x2dx=

I=∫01tan−1(2x−11+x−x2)dx=∫01tan−1(x−(1−x)1+x(1−x))dx=∫01[tan−1x−tan−1(1−x)]dx[∴ tan−1(x−y1+xy)=tan−1x−tan−1y]=∫01tan−1x−∫01tan−1(1−1-x)dx=0[∵∫0af(x)dx=∫0af(a−x)dx]

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$\int_{0}^{1} T a n^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x =$

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answer is A.

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Detailed Solution

$\begin{array}{l} I = \int_{0}^{1} \tan^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x \\ = \int_{0}^{1} \tan^{- 1} (\frac{x - (1 - x)}{1 + x (1 - x)}) d x \\ = \int_{0}^{1} [\tan^{- 1} x - \tan^{- 1} (1 - x)] d x \\ [∴ \tan^{- 1} (\frac{x - y}{1 + x y}) = \tan^{- 1} x - \tan^{- 1} y] \\ = \int_{0}^{1} \tan^{- 1} x - \int_{0}^{1} \tan^{- 1} (1 - (1 - x)) d x \\ = 0 \\ [∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] \end{array}$