The true solution set of inequality ${\log }_{(\mathrm{x}+1)}\left({\mathrm{x}}^2-4\right)>1=$

${\log }_{(\mathrm{x}+1)}\left({\mathrm{x}}^2-4\right)>1$

we have ${\mathrm{x}}^2-4>0,\mathrm{x}+1>0\text{ and }\mathrm{x}+1\ne 1$

$\therefore \mathrm{x}\in (2,\mathrm{\infty })-----\left(\mathrm{i}\right)$

Case I: 

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}+1>1\Rightarrow \mathrm{x}>0\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}^2-4>\mathrm{x}+1\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}^2-\mathrm{x}-5>0\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}\in \left(\frac{1+\sqrt{21}}{2},\mathrm{\infty }\right)---\left(\mathrm{ii}\right)\end{array}$

Case II: 

$\begin{array}{r}(\mathrm{x}+1)\in (0,1)\\ \Rightarrow \mathrm{x}\in (-1,0)\end{array}$

Clearly, this is not possible as $\mathrm{x}\in (2,\mathrm{\infty })$