1. **Problem statement:** Find the integral $$\int \ln(1+x) \, dx$$ using integration by parts. 2. **Formula and rule:** Integration by parts formula is $$\int u \, dv = uv - \int v \, du$$. 3. **Choose parts:** Let $$u = \ln(1+x)$$ and $$dv = dx$$. 4. **Compute derivatives and integrals:** - $$du = \frac{1}{1+x} dx$$ - $$v = x$$ 5. **Apply integration by parts:** $$\int \ln(1+x) \, dx = x \ln(1+x) - \int x \cdot \frac{1}{1+x} dx$$ 6. **Simplify the integral:** $$\int \frac{x}{1+x} dx = \int \frac{(1+x)-1}{1+x} dx = \int \left(1 - \frac{1}{1+x}\right) dx = \int 1 \, dx - \int \frac{1}{1+x} dx$$ 7. **Integrate each term:** $$\int 1 \, dx = x$$ $$\int \frac{1}{1+x} dx = \ln|1+x|$$ 8. **Combine results:** $$\int \frac{x}{1+x} dx = x - \ln|1+x| + C$$ 9. **Final answer:** $$\int \ln(1+x) \, dx = x \ln(1+x) - (x - \ln|1+x|) + C = (x+1) \ln(1+x) - x + C$$ This is the integral of $$\ln(1+x)$$ using integration by parts.