Question

${\int }_0^1{\tan }^{-1}\left[\frac{2x-1}{1+x-x^2}\right]dx=$

• A. 0
• B. 1/2
• C. 1
• D. $\pi /6$

Answer 1

The correct option is A 0

$I={\int }_0^{1}ta{n}^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx$
$I={\int }_0^{1}Ta{n}^{-1}\left(\frac{1-2x}{1+x-x^2}\right)dx$
$={\int }_0^{1}ta{n}^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx$
$=-{\int }_0^{1}ta{n}^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx$
$I=-I$
$\to 2I=0$
$I=0$