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

Question

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

Aicharya · Accepted Answer

$\int_{0}^{1} 	an ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}ight) d x$ $=\int_{0}^{1} 	an ^{-1}\left[\frac{x-(1-x)}{1+x(1-x)}ight] d x=\int_{0}^{1}\left[	an ^{-1} x-	an ^{-1}(1-x)ight] d x$...(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) Adding (1) \& (2), we get $2 I=0 \Rightarrow I=0$

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

Solution