Discussion Forum

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

Answer:

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