Question

The integral ${\int }_0^{1}ta{n}^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx$ simlifies to a value:

Answer 1

${\int }_0^1{\tan }^{-1}\left(\frac{2x-1}{1+x-x^2}\right)dx$
$={\int }_0^1{\tan }^{-1}\left(\frac{x+x-1}{1-x(x-1)}\right)dx$
${\int }_0^1({\tan }^{-1}x+{\tan }^{-1}(x-1\left)\right)dx$
${\int }_0^1{\tan }^{-1}xdx+{\int }_0^1{\tan }^{-1}(x-1)dx$
Applying ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$ in the first integral
$={\int }_0^1\tan {}^{-1}(1-x)dx+{\int }_0^1\tan {}^{-1}(x-1)$dx
$={\int }_0^1\tan {}^{-1}(1-x)dx-{\int }_0^1\tan {}^{-1}(1-x)dx=0$