$\int_{0}^{1} an ^{-1}\left[\frac{2 x-1}{1+x-x^{2}} ight] d x=$

Question

$\int_{0}^{1} 	an ^{-1}\left[\frac{2 x-1}{1+x-x^{2}}ight] d x=$

Aicharya · Accepted Answer

Let $\begin{aligned} I &=\int_{0}^{1} 	an ^{-1}\left[\frac{2 x-1}{1+x-x^{2}}ight] d x \ &=\int_{0}^{1} 	an ^{-1}\left[\frac{2 x-1}{1+x(1-x)}ight] d x=\int_{0}^{1} 	an ^{-1}\left[\frac{x-(1-x)}{1+x(1-x)}ight] d x \ &=\int_{0}^{1}\left[	an ^{-1} x-	an ^{-1}(1-x)ight] d x ...(1) \ &=\int_{0}^{1}\left\{	an ^{-1}(1-x)-	an ^{-1}[1-(1-x)]ight\} d x \ &=\int_{0}^{1}\left[	an ^{-1}(1-x)-	an ^{-1} xight] d x ...(2) \ 	ext { Equation }(1)+(2) \operatorname{gives} \ 2 I=0 \Rightarrow I=0 \end{aligned}$

$\int_{0}^{1} \tan ^{-1}\left[\frac{2 x-1}{1+x-x^{2}}\right] d x=$

Solution