1. **State the problem:** Solve the inequality $$\frac{x^2 + 1}{2x + 1} > 0$$ and determine the domain. 2. **Domain determination:** The denominator cannot be zero, so: $$2x + 1 \neq 0$$ $$2x \neq -1$$ $$x \neq -\frac{1}{2}$$ Thus, the domain is $$\mathbb{R} \setminus \left\{ -\frac{1}{2} \right\}$$. 3. **Analyze numerator:** The numerator is $$x^2 + 1$$. Since $$x^2 \geq 0$$ for all real $$x$$, and adding 1 makes it strictly positive: $$x^2 + 1 > 0 \quad \forall x \in \mathbb{R}$$. 4. **Analyze denominator sign:** The sign of the fraction depends on the denominator since numerator is always positive. We want: $$\frac{x^2 + 1}{2x + 1} > 0$$ Since numerator > 0, this inequality holds when denominator > 0: $$2x + 1 > 0$$ $$2x > -1$$ $$x > -\frac{1}{2}$$. 5. **Final solution:** Domain: $$\mathbb{R} \setminus \left\{ -\frac{1}{2} \right\}$$ Inequality holds for: $$x > -\frac{1}{2}$$ Therefore, the solution set is: $$\boxed{\left(-\frac{1}{2}, +\infty \right)}$$.