Answer:
Option C
Explanation:
We have,
l=$\int_{0}^{1} \tan^{-1}\left(\frac{2x-1}{1+x-x^{2}}\right)dx$
$l=\int_{0}^{1} \tan^{-1}\left(\frac{2(1-x)-1}{1+(1-x)-(1-x)^{2}}\right)dx$
$\left[ \because\int_{0}^{a} f(x)dx=\int_{0}^{a} f(a-x)dx\right]$
$l=\int_{0}^{1} \tan^{-1}\left(\frac{1-2x}{1+x-x^{2}}\right)dx$
$l=\int_{0}^{1}- \tan^{-1}\left(\frac{2x-1}{1+x-x^{2}}\right)dx$
2l=0
l=0
