$\int_{-1}^3\left(\tan ^{-1}\left(\frac{x}{x^2+1}\right)+\tan ^{-1}\left(\frac{x^2+1}{x}\right)\right) \mathrm{d} x=$

The integral is $\int_{-1}^{3} \left( \tan^{-1}\left( \frac{x}{x^{2} + 1} \right) + \tan^{-1}\left( \frac{x^{2} + 1}{x} \right) \right) \mathrm{d}x$.

Using the identity for inverse tangents, $\tan^{-1}(y) + \tan^{-1}\left( \frac{1}{y} \right) = \frac{\pi}{2}$ when $y > 0$ and $-\frac{\pi}{2}$ when $y < 0$.

Let $y = \frac{x}{x^{2} + 1}$. Since $x^{2} + 1 > 0$ for all real $x$, the sign of $y$ matches that of $x$.

For $x \in (0, 3]$, $y > 0$, so the integrand equals $\frac{\pi}{2}$.

For $x \in [-1, 0)$, $y < 0$, so the integrand equals $-\frac{\pi}{2}$.

Splitting the integral at $x = 0$ gives:
$I = \int_{-1}^{0} -\frac{\pi}{2} \mathrm{d}x + \int_{0}^{3} \frac{\pi}{2} \mathrm{d}x = -\frac{\pi}{2}(0 - (-1)) + \frac{\pi}{2}(3 - 0) = -\frac{\pi}{2} + \frac{3\pi}{2} = \pi$.

The final result is $\pi$.