Where am I wrong in $\int_0^1 x(1-x)^ndx$ We have $\int_0^1 x(1-x)^ndx=-\frac{x(1-x)^{n+1}}{n+1}|^{^1}_{_0}+\frac{(1-x)^{n+2}}{(n+1)(n+2)}|^{^1}_{_0}=-\frac{1}{(n+1)(n+2)}$ 
If I use substitution $u=1-x\Rightarrow \int_0^1 (1-u)u^n du=\frac{u^{n+1}}{n+1}|^{^1}_{_0}-\frac{u^{n+2}}{n+2}|^{^1}_{_0}=\frac{1}{(n+1)(n+2)}$

Where am I wrong ?


REVISED BY REQUEST
We begin with the integral $\int_0^1 x(1-x)^ndx$.  Integrating by parts gives
$$\begin{align}
\int_0^1 x(1-x)^ndx&=\left.-\frac{x(1-x)^{n+1}}{n+1}\right|^{^1}_{_0}+\frac{1}{n+1}\int_0^1 (1-x)^{n+1}dx\\\\
&=\frac{1}{n+1}\int_0^1 (1-x)^{n+1}dx
\end{align}$$
Now, integrating the remaining term reveals that
$$\begin{align}
\frac{1}{n+1}\int_0^1 (1-x)^{n+1}dx&=-\frac{1}{n+1}\left.\frac{(1-x)^{n+2}}{n+2}\right|^{^1}_{_0}\\\\
&=\frac{1}{(n+1)(n+2)}
\end{align}$$

ORIGINAL ANSWER
$$\begin{align}
\int_0^1 x(1-x)^ndx&=\left.-\frac{x(1-x)^{n+1}}{n+1}\right|^{^1}_{_0}\color{red}{-}\left.\frac{(1-x)^{n+2}}{(n+1)(n+2)}\right|^{^1}_{_0}\\\\
&=\frac{1}{(n+1)(n+2)}
\end{align}$$
