If $\left(\tan ^{-1} x\right)^2+\left(\cot ^{-1} x\right)^2=\frac{5 \pi^2}{8}$, then $x^2+1=$

Let $A = \tan^{-1} x$. Using the identity $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$, we have $\cot^{-1} x = \frac{\pi}{2} - A$.

Substituting into the equation $(\tan^{-1} x)^2 + (\cot^{-1} x)^2 = \frac{5\pi^2}{8}$ yields:

$A^2 + \left(\frac{\pi}{2} - A\right)^2 = \frac{5\pi^2}{8}$

Expanding gives:

$A^2 + \frac{\pi^2}{4} - \pi A + A^2 = \frac{5\pi^2}{8}$

Simplifying and combining terms:

$2A^2 - \pi A + \frac{2\pi^2 - 5\pi^2}{8} = 0$
$2A^2 - \pi A - \frac{3\pi^2}{8} = 0$

Multiplying through by 8:

$16A^2 - 8\pi A - 3\pi^2 = 0$

Solving this quadratic using the formula:

$A = \frac{8\pi \pm \sqrt{64\pi^2 + 192\pi^2}}{32}$
$A = \frac{8\pi \pm 16\pi}{32}$

Yielding solutions $A = \frac{3\pi}{4}$ and $A = -\frac{\pi}{4}$.

Since the principal range of $\tan^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$, only $A = -\frac{\pi}{4}$ is valid.

Thus $x = \tan(-\frac{\pi}{4}) = -1$, and:

$x^2 + 1 = (-1)^2 + 1 = 2$