1. **State the problem:** Solve the inequality $$\frac{x^2 - 3x - 4}{x + 1} < 0$$. 2. **Factor the numerator:** The quadratic $x^2 - 3x - 4$ factors as $$(x - 4)(x + 1)$$. 3. **Rewrite the inequality:** Substitute the factorization into the inequality: $$\frac{(x - 4)(x + 1)}{x + 1} < 0$$. 4. **Simplify the expression:** For $x \neq -1$, the $(x + 1)$ terms cancel out, so the inequality reduces to: $$x - 4 < 0$$. 5. **Consider the domain:** The original expression is undefined at $x = -1$ because the denominator is zero there. 6. **Solve the simplified inequality:** $$x - 4 < 0 \implies x < 4$$. 7. **Analyze the sign around $x = -1$:** Since the original expression is undefined at $x = -1$, we must exclude this point. 8. **Check the sign of the original expression on intervals:** - For $x < -1$, numerator $(x - 4)(x + 1)$ is positive times negative = negative, denominator $x + 1$ is negative, so the whole fraction is positive (negative/negative = positive). - For $-1 < x < 4$, numerator is negative times positive = negative, denominator positive, so fraction is negative. - For $x > 4$, numerator positive, denominator positive, fraction positive. 9. **Conclusion:** The inequality holds where the fraction is less than zero, which is for $$-1 < x < 4$$. 10. **Final answer:** $$\boxed{(-1, 4)}$$. This means all $x$ values strictly between $-1$ and $4$ satisfy the inequality.