Answer
$(-2, -1)\cup(2,\infty)$
Work Step by Step
Given $\frac{x^2+3x-1}{x+1}\gt3$, we can identify it as an inequality. Graph $f(x)=\frac{x^2+3x-1}{x+1}-3=\frac{x^2-4}{x+1}$ as shown in the figure and we can find the solution set as $(-2, -1)\cup(2,\infty)$