The sum of possible values of $x$ for ${\tan }^{-1}\left(x+1\right)+{\cot }^{-1}\left(\frac{1}{x-1}\right)={\tan }^{-1}\left(\frac{8}{31}\right)$ is:

Given ${\tan }^{-1}\left(x+1\right)+{\cot }^{-1}\left(\frac{1}{x-1}\right)={\tan }^{-1}\frac{8}{31}$

$\Rightarrow {\tan }^{-1}\left(x+1\right)+{\tan }^{-1}\left(x-1\right)={\tan }^{-1}\frac{8}{31}, \left(\therefore {\tan }^{-1}x={\cot }^{-1}\frac{1}{x}, x>0\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{x+1+x-1}{1-\left(x+1\right)\left(x-1\right)}\right)={\tan }^{-1}\frac{8}{31}$

$\Rightarrow {\tan }^{-1}\left[\frac{2x}{1-\left(x+1\right)\left(x-1\right)}\right]={\tan }^{-1}\frac{8}{31}$

Taking tangent both the sides we get, 

$\frac{(x+1)+(x-1)}{1-\left(x^2-1\right)}=\frac{8}{31}$

$\Rightarrow \frac{2x}{2-x^2}=\frac{8}{31}$

$\Rightarrow 4x^2+31x-8=0$

$\Rightarrow x=-8,\frac{1}{4}$

But, if $x=\frac{1}{4}$

${\tan }^{-1}\left(x+1\right)\in \left(0,\frac{\pi }{2}\right)$

and ${\cot }^{-1}\left(\frac{1}{x-1}\right)\in \left(\frac{\pi }{2},\pi \right)$

$\Rightarrow \mathrm{LHS}>\frac{\pi }{2}$ and $\mathrm{RHS}<\frac{\pi }{2}$ (Not possible)

Hence, $x=-8=-\frac{32}{4}.$