$9.252 \log _{x}\left(x^{3}+1\right) \cdot \log _{x+1} x>2$.

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$9.252 \log _{x}\left(x^{3}+1\right) \cdot \log _{x+1} x>2$.
Solution. $\log _{x}\left((x+1)\left(x^{2}-x+1\right)\right) \cdot \log _{x+1} x>2 \Leftrightarrow$

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

$\Leftrightarrow \frac{\log _{x}\left(x^{2}-x+1\right)}{\log _{x}(x+1)}>1 \Leftrightarrow\left\{\begin{array}{l}\log _{x+1}\left(x^{2}-x+1\right)>1, \\ x>0 ; x \neq 1\end{array} \Leftrightarrow\left\{\begin{array}{l}x^{2}-x+1>x+1, \\ x>0 ; x \neq 1\end{array} \Leftrightarrow x>2\right.\right.$.

Answer: $x \in(2 ;+\infty)$.