Question

Obtain: ${\int }_0^1{\tan }^{-1}\left(\frac{1}{1-x+x^2}\right)dx$, where $0<x<1$

Answer 1

$I={\int }_0^1{\tan }^{-1}\left(\frac{1}{1-x+x^2}\right)dx$

$I={\int }_0^1{\tan }^{-1}\left(\frac{x+(1-x)}{1-x(1-x}\right)dx$

$I={\int }_0^1[{\tan }^{-1}x+{\tan }^{-1}(1-x)]dx$

We know,

${\int }_0^1{\tan }^{-1}\left(x\right)dx={\int }_0^1{\tan }^{-1}(1-x)dx$

$\therefore I=2{\int }_0^1{\tan }^{-1}xdx$

$\therefore$ Integrating by parts:

$\therefore I=2{\left[x{\tan }^{-1}\right(x)-\frac{1}{2}ln(1+x^2\left)\right]}_0^1$

$\therefore I=[\frac{\pi }{4}-ln(\sqrt{2}\left)\right]$