1. **Problem:** Evaluate the integral $$\int \sqrt{x+1} \, dx$$ using substitution. 2. **Step 1: Substitution** Let $$u = x + 1$$, then $$du = dx$$. 3. **Step 2: Rewrite the integral** The integral becomes $$\int \sqrt{u} \, du = \int u^{1/2} \, du$$. 4. **Step 3: Integrate** Use the power rule for integration: $$\int u^n \, du = \frac{u^{n+1}}{n+1} + C$$. So, $$\int u^{1/2} \, du = \frac{u^{3/2}}{3/2} + C = \frac{2}{3} u^{3/2} + C$$. 5. **Step 4: Substitute back** Replace $$u$$ with $$x+1$$: $$\frac{2}{3} (x+1)^{3/2} + C$$. 6. **Step 5: Verification by differentiation** Differentiate $$F(x) = \frac{2}{3} (x+1)^{3/2} + C$$: $$F'(x) = \frac{2}{3} \cdot \frac{3}{2} (x+1)^{1/2} \cdot 1 = \sqrt{x+1}$$, which matches the integrand. **Final answer:** $$\int \sqrt{x+1} \, dx = \frac{2}{3} (x+1)^{3/2} + C$$