Indefinte integration reloaded

akki ~~ unlucky forever ~~

$I=\int_{0}^{1}tan^{-1}\frac{1}{(1-x+x^2)}dx\\ \\ I=\int_{0}^{1}tan^{-1}(\frac{(1-x )+ x}{1-x(1-x)})dx\\ \\ I=\int_{0}^{1}tan^{-1}(1-x)\ dx+\int_{0}^{1}tan^{-1}x\ dx\\ \\ applying\ integration\ b\ parts\ to\ each\\ \\ I=\int_{0}^{1}\frac{1-x}{1+x^2}dx+\frac{\pi}{4}-\int_{0}^{1}\frac{x\ dx}{1+x^2}\\ \\ I=\frac{\pi}{2}-\int_{0}^{1}\frac{2x\ dx}{1+x^2}\\ \\ I=\frac{\pi}{2}-ln2$

.