1. The problem is to evaluate the definite integral $$\int_1^{e-1} \ln(n+1) \, dn$$. 2. We use integration by parts formula: $$\int u \, dv = uv - \int v \, du$$. 3. Let $$u = \ln(n+1)$$ and $$dv = dn$$. 4. Then, $$du = \frac{1}{n+1} dn$$ and $$v = n$$. 5. Applying integration by parts: $$\int \ln(n+1) \, dn = n \ln(n+1) - \int \frac{n}{n+1} dn$$. 6. Simplify the integral $$\int \frac{n}{n+1} dn$$: Rewrite $$\frac{n}{n+1} = \frac{n+1-1}{n+1} = 1 - \frac{1}{n+1}$$. 7. So, $$\int \frac{n}{n+1} dn = \int 1 \, dn - \int \frac{1}{n+1} dn = n - \ln|n+1| + C$$. 8. Substitute back: $$\int \ln(n+1) \, dn = n \ln(n+1) - (n - \ln|n+1|) + C = n \ln(n+1) - n + \ln|n+1| + C$$. 9. Evaluate the definite integral from 1 to $$e-1$$: $$\left[ n \ln(n+1) - n + \ln(n+1) \right]_1^{e-1}$$. 10. Calculate at upper limit $$n = e-1$$: $$ (e-1) \ln(e) - (e-1) + \ln(e) = (e-1)(1) - (e-1) + 1 = (e-1) - (e-1) + 1 = 1$$. 11. Calculate at lower limit $$n=1$$: $$1 \ln(2) - 1 + \ln(2) = \ln(2) - 1 + \ln(2) = 2 \ln(2) - 1$$. 12. Subtract lower from upper: $$1 - (2 \ln(2) - 1) = 1 - 2 \ln(2) + 1 = 2 - 2 \ln(2) = 2(1 - \ln(2))$$. 13. Final answer: $$\int_1^{e-1} \ln(n+1) \, dn = 2(1 - \ln(2))$$.