Solution

Let $t=1-x \Rightarrow x=1-t,\ dx=-dt$. 

As $x:0\to1$, 

$t:1\to0$.

\[ \int_{0}^{1} x(1-x)^n\,dx \] 

\[       = \int_{1}^{0} (1-t)\,t^n\,(-dt) \] 

\[= \int_{0}^{1} \big(t^n - t^{n+1}\big)\,dt \] 

\[       = \left[\frac{t^{n+1}}{n+1}-\frac{t^{n+2}}{n+2}\right]_{0}^{1} = \frac{1}{n+1}-\frac{1}{n+2}. \]

Simplify: \[ \frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}. \]

Answer: $\boxed{\dfrac{1}{(n+1)(n+2)}}$ (valid for $n>-1$).