Answer
$$\frac{1}{{{n^2} + 3n + 2}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {x{{\left( {1 - x} \right)}^n}dx} \cr
& {\text{Let }}u = 1 - x \Rightarrow x = 1 - u,{\text{ }}dx = - du \cr
& {\text{The new limits of integration are:}} \cr
& x = 1 \Rightarrow u = 0 \cr
& x = 0 \Rightarrow u = 1 \cr
& {\text{Apply the substitution}} \cr
& \int_0^1 {x{{\left( {1 - x} \right)}^n}dx} = \int_1^0 {\left( {1 - u} \right){u^n}\left( { - du} \right)} \cr
& {\text{ }} = \int_0^1 {\left( {1 - u} \right){u^n}du} \cr
& {\text{ }} = \int_0^1 {\left( {{u^n} - {u^{n + 1}}} \right)du} \cr
& {\text{Integrating}} \cr
& {\text{ }} = \left[ {\frac{{{u^{n + 1}}}}{{n + 1}} - \frac{{{u^{n + 2}}}}{{n + 2}}} \right]_0^1 \cr
& {\text{Evaluating}} \cr
& {\text{ }} = \left[ {\frac{{{1^{n + 1}}}}{{n + 1}} - \frac{{{1^{n + 2}}}}{{n + 2}}} \right] - \left[ {\frac{{{0^{n + 1}}}}{{n + 1}} - \frac{{{0^{n + 2}}}}{{n + 2}}} \right] \cr
& {\text{ }} = \frac{1}{{n + 1}} - \frac{1}{{n + 2}} \cr
& {\text{ }} = \frac{{n + 2 - n - 1}}{{\left( {n + 1} \right)\left( {n + 2} \right)}} \cr
& {\text{ }} = \frac{1}{{{n^2} + 3n + 2}} \cr} $$
