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

$\begin{array}{l}\text{The given eqution is }{\tan }^{-1}\left(x+1\right)+{\cot }^{-1}\left(\frac{1}{x-1}\right)={\tan }^{-1}\left(\frac{8}{31}\right)\\ \text{it implies }{\tan }^{-1}\left(x+1\right)+{\tan }^{-1}\left(x-1\right)={\tan }^{-1}\left(\frac{8}{31}\right)\\ \text{Hence, }\Rightarrow {\tan }^{-1}\left(\frac{x+1+x-1}{1-\left(x+1\right)\left(x-1\right)}\right)={\tan }^{-1}\left(\frac{8}{31}\right)\\ \Rightarrow \frac{2x}{2-x^2}=\frac{8}{31}\\ \Rightarrow 31x=8-4x^2\\ \Rightarrow 4x^2+31x-8=0\\ x=-8,\frac{1}{4}\\ \text{by substituting the values of }x\text{\hspace{0.17em}in the given equation, conclude that }x=-8\text{\hspace{0.17em}only the solution }\end{array}$