Evaluating the integral $\int_0^1\arctan(1-x+x^2)dx$

$\int_0^1\tan^{-1}(1-x+x^2)dx=\frac{\pi}{2}-\int_{0}^{1}\tan^{-1}\frac1{1-x+x^2}dx=\frac{\pi}{2}-\int_0^1 \tan^{-1}(x)-\tan^{-1}(x-1)dx$ That should simplify everything.