1. **State the problem:** We need to evaluate the definite integral $$\int_0^2 \ln(1+x) \, dx$$ which represents the area under the curve $y = \ln(1+x)$ from $x=0$ to $x=2$. 2. **Formula and method:** To solve this integral, we use integration by parts. Recall the formula: $$\int u \, dv = uv - \int v \, du$$ 3. **Choose parts:** Let $$u = \ln(1+x) \quad \Rightarrow \quad du = \frac{1}{1+x} dx$$ $$dv = dx \quad \Rightarrow \quad v = x$$ 4. **Apply integration by parts:** $$\int_0^2 \ln(1+x) \, dx = \left. x \ln(1+x) \right|_0^2 - \int_0^2 \frac{x}{1+x} dx$$ 5. **Evaluate the boundary term:** $$\left. x \ln(1+x) \right|_0^2 = 2 \ln(3) - 0 = 2 \ln(3)$$ 6. **Simplify the remaining integral:** $$\int_0^2 \frac{x}{1+x} dx = \int_0^2 \frac{(1+x)-1}{1+x} dx = \int_0^2 \left(1 - \frac{1}{1+x}\right) dx$$ 7. **Split the integral:** $$= \int_0^2 1 \, dx - \int_0^2 \frac{1}{1+x} dx = \left. x \right|_0^2 - \left. \ln|1+x| \right|_0^2 = 2 - \ln(3)$$ 8. **Combine all parts:** $$\int_0^2 \ln(1+x) \, dx = 2 \ln(3) - (2 - \ln(3)) = 2 \ln(3) - 2 + \ln(3) = 3 \ln(3) - 2$$ 9. **Final answer:** $$\boxed{3 \ln(3) - 2}$$ This is the exact value of the integral.