If 2 ∫ 0 1 tan - 1 x d x = ∫ 0 1 cot - 1 1 - x + x 2 d x , then ∫ 0 1 tan - 1 1 - x + x 2…

From the given equation

$2\underset{0}{\overset{1}{\int }}{\tan }^{-1}xdx= \underset{0}{\overset{1}{\int }}\left(\frac{\pi }{2}-{\tan }^{-1}\left(1-x+x^2\right)\right)dx$

$\Rightarrow 2\underset{0}{\overset{1}{\int }}{\tan }^{-1}xdx= \underset{0}{\overset{1}{\int }}\frac{\pi }{2} dx- \underset{0}{\overset{1}{\int }}{\tan }^{-1}\left(1-x+x^2\right)dx$
 $\Rightarrow \underset{0}{\overset{1}{\int }}{\tan }^{-1}\left(1-x+x^2\right)dx=\frac{\pi }{2}-2 \underset{0}{\overset{1}{\int }}{\tan }^{-1}xdx \dots \left(\mathrm{i}\right)$         
Let $I_1= \underset{0}{\overset{1}{\int }}{\tan }^{-1}xdx$

$={\left[\left({\tan }^{-1}x\right)x\right]}_0^1-\underset{0}{\overset{1}{\int }}\frac{1}{1+x^2}xdx$

$=\frac{\pi }{4}- \underset{0}{\overset{1}{\int }}\frac{x}{1+x^2}dx$

$=\frac{\pi }{4}-\frac{1}{2} \underset{0}{\overset{1}{\int }}\frac{2x}{1+x^2}dx$

$=\frac{\pi }{4}-\frac{1}{2} {\left[\ln \left(1+x^2\right)\right]}_0^1$

$=\frac{\pi }{4}-\frac{1}{2}\ln 2$

By equation $\left(\mathrm{i}\right)$,

${\int }_0^1{\tan }^{-1}\left(1-x+x^2\right)dx=$ $\frac{\pi }{2}-2\left[\frac{\pi }{4}-\frac{1}{2}\ln 2\right]=\ln 2$