Find the value of $\int\limits_{0}^{1} \tan^{-1} \left( \frac{1-2x}{1+x-x^2} \right) dx$.
Answer & explanation
Correct answer: option 3
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$
$= \int\limits_{0}^{1} \tan^{-1}(1-x) - \tan^{-1}x dx$
$=0$
$\left[\tan^{-1}x - \tan^{-1}y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)\right]$.
$\left[\text{0 as}\int\limits_{0}^{a} \tan^{-1}x\, dx = \int\limits_{0}^{a} \tan^{-1}(1-x) dx\right]$