A) 1
B) 0
C) \[-1\]
D) \[\frac{\pi }{4}\]
Correct Answer: B
Solution :
Let \[l=\int\limits_{0}^{1}{{{\tan }^{-1}}\left( \frac{2x-1}{1+x-{{x}^{2}}} \right)}dx\] \[=\int\limits_{0}^{1}{{{\tan }^{-1}}\left\{ \frac{x+(x-1)}{1-x(x-1)} \right\}}dx\] \[=\int\limits_{0}^{1}{\{{{\tan }^{-1}}x+{{\tan }^{-1}}(x-1)\}}dx\] \[\left\{ \because \,{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right) \right\}\] \[\Rightarrow \] \[l=\int\limits_{0}^{1}{\{{{\tan }^{-1}}x-{{\tan }^{-1}}(1-x)\}}dx\] ?(i) Also \[l=\int\limits_{0}^{1}{\{{{\tan }^{-1}}(1-x)-{{\tan }^{-1}}\{1-(1-x)\}\}}dx\] \[\left[ \because \,\int_{0}^{a}{f(x)dx=}\int_{0}^{a}{f(a-x)dx} \right]\] \[l=\int\limits_{0}^{1}{[{{\tan }^{-1}}(1-x)-{{\tan }^{-1}}x]}dx\] ?(ii) On adding Eqs.(i) and (ii), we get \[2l=0\] \[\Rightarrow \] \[l=0\]
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