\[f(x) = \frac{x^{2} + 7}{x + 1};\ \ x \neq - 1\]
\[\frac{x^{2} + 7}{x + 1} = 4^{\backslash x + 1}\]
\[x^{2} + 7 = 4x + 1\]
\[x^{2} + 7x - 4x - 4 = 0\]
\[x^{2} - 4x + 3 = 0\]
\[x_{1} + x_{2} = 4;\ \ \ x_{1} \cdot x_{2} = 3\]
\[x_{1} = 1;\ \ \ \ \ \ \ \ \ \ \ \ x_{2} = 3.\]
\[Ответ:при\ x = 1;x = 3.\]