1. **State the problem:** We want to find the integral $$\int \ln(x+1) \, dx$$ using integration by parts. 2. **Recall integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$. 3. **Choose parts:** Let $$u = \ln(x+1)$$ and $$dv = dx$$. 4. **Compute derivatives and integrals:** - $$du = \frac{1}{x+1} dx$$ - $$v = x$$ 5. **Apply integration by parts:** $$\int \ln(x+1) \, dx = x \ln(x+1) - \int x \cdot \frac{1}{x+1} dx$$ 6. **Simplify the integral:** $$\int \frac{x}{x+1} dx = \int \frac{x+1-1}{x+1} dx = \int \left(1 - \frac{1}{x+1}\right) dx = \int 1 \, dx - \int \frac{1}{x+1} dx$$ 7. **Integrate each term:** $$\int 1 \, dx = x$$ $$\int \frac{1}{x+1} dx = \ln|x+1|$$ 8. **Combine results:** $$\int \frac{x}{x+1} dx = x - \ln|x+1| + C$$ 9. **Substitute back:** $$\int \ln(x+1) \, dx = x \ln(x+1) - (x - \ln|x+1|) + C = x \ln(x+1) - x + \ln|x+1| + C$$ 10. **Final answer:** $$\boxed{\int \ln(x+1) \, dx = (x+1) \ln(x+1) - x + C}$$