1. **State the problem:** Find the domain of the function $$f(u) = \frac{u+1}{1 + \frac{1}{u+1}}$$. 2. **Recall the domain rules:** The domain of a function includes all values of $u$ for which the function is defined. Here, the function involves division, so the denominator cannot be zero. Also, the expression inside the denominator has a fraction with $u+1$ in the denominator, so $u+1 \neq 0$. 3. **Identify restrictions:** - From the inner denominator: $$u+1 \neq 0 \implies u \neq -1$$ - From the overall denominator: $$1 + \frac{1}{u+1} \neq 0$$ 4. **Simplify the denominator:** $$1 + \frac{1}{u+1} = \frac{(u+1) + 1}{u+1} = \frac{u+2}{u+1}$$ 5. **Set denominator not equal to zero:** $$\frac{u+2}{u+1} \neq 0$$ 6. **A fraction is zero only if numerator is zero, so denominator is zero if numerator is zero:** $$u+2 \neq 0 \implies u \neq -2$$ 7. **Combine restrictions:** $$u \neq -1 \text{ and } u \neq -2$$ 8. **Conclusion:** The domain of $f(u)$ is all real numbers except $u = -1$ and $u = -2$. **Final answer:** $$\boxed{\{u \in \mathbb{R} : u \neq -1, u \neq -2\}}$$