The value of $\int_0^1 \tan ^{-1}\left(1-x+x^2\right) \mathrm{d} x$ is

The integral $I = \int_0^1 \tan^{-1}(1-x+x^2) dx$ can be evaluated using trigonometric identities and substitution techniques.

Since $1-x+x^2 = \left(x-\frac{1}{2}\right)^2 + \frac{3}{4} > 0$ for $x \in [0,1]$, the identity $\tan^{-1} A + \tan^{-1} \frac{1}{A} = \frac{\pi}{2}$ applies, giving $\tan^{-1}(1-x+x^2) = \frac{\pi}{2} - \tan^{-1}\left(\frac{1}{1-x+x^2}\right)$.

Substituting yields $I = \int_0^1 \frac{\pi}{2} dx - \int_0^1 \tan^{-1}\left(\frac{1}{1-x+x^2}\right) dx = \frac{\pi}{2} - \int_0^1 \tan^{-1}\left(\frac{1}{1-x+x^2}\right) dx$.

Using the identity $\tan^{-1} x - \tan^{-1} (x-1) = \tan^{-1}\left(\frac{1}{1-x+x^2}\right)$, the integral becomes $I = \frac{\pi}{2} - \int_0^1 (\tan^{-1} x - \tan^{-1} (x-1)) dx$.

Substituting $u = x-1$ in the second term gives $\int_0^1 \tan^{-1} (x-1) dx = \int_{-1}^0 \tan^{-1} u du = -\int_0^1 \tan^{-1} u du$ by odd symmetry.

Thus, $I = \frac{\pi}{2} - \left(\int_0^1 \tan^{-1} x dx + \int_0^1 \tan^{-1} x dx\right) = \frac{\pi}{2} - 2\int_0^1 \tan^{-1} x dx$.

Integration by parts with $v = \tan^{-1} x$, $du = dx$ yields $\int_0^1 \tan^{-1} x dx = [x \tan^{-1} x]_0^1 - \int_0^1 \frac{x}{1+x^2} dx = \frac{\pi}{4} - \int_0^1 \frac{x}{1+x^2} dx$.

The remaining integral evaluates to $\frac{1}{2}\log 2$ via substitution $t = 1+x^2$, giving $\int_0^1 \tan^{-1} x dx = \frac{\pi}{4} - \frac{1}{2}\log 2$.

Substituting back, $I = \frac{\pi}{2} - 2\left(\frac{\pi}{4} - \frac{1}{2}\log 2\right) = \log 2$.

Final answer: $\log 2$