Find the value of $\int\limits_{0}^{1} \tan^{-1} \left( \frac{1-2x}{1+x-x^2} \right) dx$.

$0$

The correct answer is Option (3) → 0

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

$ = \int\limits_{0}^{1} \tan^{-1} \left( \frac{(1-x)-x}{1+(1-x)x} \right) dx$

$= \int\limits_{0}^{1}  \tan^{-1}(1-x) - \tan^{-1}x  dx$

$=0$

$\left[\tan^{-1}x - \tan^{-1}y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)\right]$.

$\left[\text{0 as}\int\limits_{0}^{a} \tan^{-1}x\, dx = \int\limits_{0}^{a} \tan^{-1}(1-x) dx\right]$