The value of $\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x$ is
A. 1
B. 0
C. $-1$
D. $\frac{\pi}{4}$
Let $I=\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x$
$\Rightarrow I=\int_{0}^{1} \tan ^{-1}\left(\frac{x-(1-x)}{1+x(1-x)}\right) d x$
$\Rightarrow I=\int_{0}^{1}\left[\tan ^{-1} x-\tan ^{-1}(1-x)\right] d x$ ...(1)
$\Rightarrow I=\int_{0}^{1}\left[\tan ^{-1}(1-x)-\tan ^{-1}(1-1+x)\right] d x$
$\Rightarrow I=\int_{0}^{1}\left[\tan ^{-1}(1-x)-\tan ^{-1}(x)\right] d x$
$\Rightarrow I=\int_{0}^{1}\left[\tan ^{-1}(1-x)-\tan ^{-1}(x)\right] d x$ ...(2)
Adding (1) and (2), we obtain
$2 I=\int_{0}^{1}\left(\tan ^{-1} x+\tan ^{-1}(1-x)-\tan ^{-1}(1-x)-\tan ^{-1} x\right) d x$
$\Rightarrow 2 I=0$
$\Rightarrow I=0$
Hence, the correct answer is B.