Exponential inequality

$\begin{array}{l} {\left( {\frac{1}{3}} \right)^{{x^2} + 1}} > {\left( {\frac{1}{9}} \right)^{x + 2}}\\ \Leftrightarrow {\left( {{3^{ - 1}}} \right)^{{x^2} + 1}} > {\left( {{3^{ - 2}}} \right)^{x + 2}}\\ \Leftrightarrow {\left( 3 \right)^{ - {x^2} - 1}} > {\left( 3 \right)^{ - 2x - 4}}\\ \Leftrightarrow - {x^2} - 1 > - 2x - 4\\ \Leftrightarrow - {x^2} + 2x + 3 > 0\\ \Leftrightarrow {x^2} - 2x - 3 < 0\\ \Leftrightarrow - 1 < x < 3\\ \\ \to \beta - \alpha = 3 - ( - 1) = 4 \end{array}$

the answer is wrong ! the solution to this equation would be : x belongs to (- infinity to -1) U ( 3 to +infinity)