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 ?

 A: Your first sentence is not correct : 
$$\begin{align}\int_{0}^{1}x(1-x)^ndx&=\int_{0}^{1}x\left(\frac{-(1-x)^{n+1}}{n+1}\right)'dx\\&=\left[x\left(\frac{-(1-x)^{n+1}}{n+1}\right)\right]_{0}^{1}-\int_{0}^{1}\frac{-(1-x)^{n+1}}{n+1}dx\\&=\left[x\left(\frac{-(1-x)^{n+1}}{n+1}\right)\right]_{0}^{1}\color{red}{-}\left[\frac{(1-x)^{n+2}}{(n+1)(n+2)}\right]_{0}^{1}\end{align}$$
A: 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
$$\begin{align}
\int_0^1 x(1-x)^ndx&=\left.-\frac{x(1-x)^{n+1}}{n+1}\right|^{^1}_{_0}\color{red}{-}\left.\frac{(1-x)^{n+2}}{(n+1)(n+2)}\right|^{^1}_{_0}\\\\
&=\frac{1}{(n+1)(n+2)}
\end{align}$$
A: It can be done by a simple move: Let $u = 1-x$, then the integral becomes a much simpler one: $I = \displaystyle \int_{0}^1 (u^n-u^{n+1})du$. And you can take it from here...
