# 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.

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## 1 Answer

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

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why would you think in that way?Is there any hint motivated us to do that? –  Vulcan Dec 7 '12 at 12:43
The reduction fromulae you got will not help you reach your goal, thus its time to try something else –  Amr Dec 7 '12 at 12:45