1. We need to evaluate $I=\int_0^1 x(1-x)^n\,dx.$

2. Use the substitution $u=1-x \implies x=1-u, \quad dx=-du.$ When $x=0$, $u=1$; when $x=1$, $u=0$.

So, $I=\int_1^0 (1-u)u^n(-du)=\int_0^1 (1-u)u^n\,du.$

3. Expand the integrand: $I=\int_0^1 (u^n-u^{n+1})\,du.$

4. Integrate term-by-term: $I=\int_0^1 u^n\,du-\int_0^1 u^{n+1}\,du.$ Using $\int_0^1 u^m\,du=\frac{1}{m+1},$ we get $I=\frac{1}{n+1}-\frac{1}{n+2}.$

5. Compare with the options:

• A: $\frac{1}{n+1}+\frac{1}{n+2}$ ❌
• B: $\frac{1}{n+1}$ ❌
• C: $\frac{1}{n+2}$ ❌
• D: $\frac{1}{n+1}-\frac{1}{n+2}$ ✅

Therefore, the correct option is $\boxed{D}.$