Reduction Formula $ I_{(m,n)}\; =\;\int x^m(x+a)^ndx $ Some problem occured in proving the following reduction formula.
$$
\\
I_{(m,n)}\; =\;\int x^m(x+a)^ndx\; = \; \frac{x^m(x+a)^{n+1}}{m+n+1}-\frac{ma}{m+n+1}I_{(m-1,n)}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;m,n \in N
$$
I have tried by using integration by part,here are my result
$$
\begin{align}
I_{(m,n)}\; =\;\frac{x^{m+1}(x+a)^n}{m+1}-\frac{n}{m+1}I_{(m+1,n-1)}\\
I_{(m,n)}\; =\;\frac{x^{m}(x+a)^{n+1}}{n+1}-\frac{m}{n+1}I_{(m-1,n+1)}
\end{align}
$$
I have no idea on how to combine the 2 result or my direction of attacking the problem is wrong.
Any help would be appreciated.Thank you.

Hint:
$I(m,n)=\int x^m(x+a)^ndx=\int x^{m-1}(x+a-a)(x+a)^{n}dx=\int x^{m-1}(x+a)^{n+1}dx-a\int x^{m-1}(x+a)^{n}dx=I(m-1,n+1)-aI(m-1,n)$
Use this with your second reduction formula
