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Fundamental theorem of calculus - class-XII
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The value of $\displaystyle \int _0^1\tan^{-1}\left (\frac {2x-1}{1+x-x^2}\right )dx$ is
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A
$0$
💡 Explanation:
Let $I=\int _0^1\tan^{-1}\left (\dfrac {2x-1}{1+x-x^2}\right )dx$
$\Rightarrow I=\int _0^1 \tan^{-1}\left (\dfrac {x-(1-x)}{1+x(1-x)}\right )dx$
$\Rightarrow I=\int _0^1[\tan^{-1}x-\tan^{-1}(1-x)]dx$ ................ (1)
$\Rightarrow I=\int _0^1[\tan^{-1}(1-x)-\tan^{-1}(1-1+x)]dx$
$\Rightarrow I=\int _0^1[\tan^{-1}(1-x)-\tan^{-1}(x)]dx$
$\Rightarrow I=\int _0^1[\tan^{-1}(1-x)-\tan^{-1}(x)]dx$ ........... (2)
Adding (1) and (2), we obtain
$2I=\int _0^1(\tan^{-1}x+\tan^{-1}(1-x)-\tan^{-1}(1-x)-\tan^{-1}x)dx=0$
$\Rightarrow I=0$
Hence, the correct Answer is B.