$\int\limits_0^1\tan^{-1}(\frac{2x-1}{1+x-x^2})dx$ is equal to

0

The correct answer is Option (2) → 0

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

Put $I=\int_{0}^{1}\tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right)\,dx$

Use property

$\tan^{-1}u+\tan^{-1}\frac{1}{u}=\frac{\pi}{2}$ for $u>0$

Change variable $x=1-t$

$I=\int_{0}^{1}\tan^{-1}\left(\frac{1-2t}{1+t-t^2}\right)\,dt$

Add both expressions

$2I=\int_{0}^{1}\left[\tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right)+\tan^{-1}\left(\frac{1-2x}{1+x-x^2}\right)\right]dx$

$=0$

$I=0$

The value of the given integral is $0$.