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Question:

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. $\frac{\pi}{4}$

Solution:

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.

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