[tex]\log_{\dfrac{x+1}{x^2+1}}(4-x) = \dfrac{\ln(4-x)}{\ln\Big(\dfrac{x+1}{x^2+1}\Big)}\\ \\ \Rightarrow \begin{cases} 4-x >0\\\\ \dfrac{x+1}{x^2+1}>0\\ \\\dfrac{x+1}{x^2+1}\neq 1\end{cases} \Rightarrow \begin{cases} x<4\\\\ x>-1,~~(\text{fiindca }x^2+1 > 0,\forall x\in \mathbb{R})\\ \\x\neq \{0,1\}\end{cases} \Rightarrow \\ \\\\ \Rightarrow x\in (-1,0)\cup(0,1)\cup(1,4)[/tex]
