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Quoted from: Finding a reduction formula for this integral.

However, I am not sure how to arrive at this: $$\begin{align} \phantom{=}&\color{green}{\int_0^1x^{n+2}(1-x)^n\,dx}+2\int_0^1x^{n+1}(1-x)^{n+1}\,dx+\color{green}{\int_0^1x^n(1-x)^{n+2}\,dx}\\ =&2I(n+1)+\color{green}{2\int_0^1x^{n+2}(1-x)^n\,dx} \end{align}$$

and why the highlighted part is equal to each other? What type of integral is that?

metamorphy
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Max Su
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    Symmetry about $x=\frac12$ – Henry Jul 16 '21 at 16:22
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    Hi, Welcome to Stack Exchange! I have edited your question for you but in the future please use Mathjax (similar to LaTeX) to write any questions as it is more accessible for other users. Here is a tutorial: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Henry Lee Jul 16 '21 at 16:36

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Just to clarify, if symmetry is no the first thing you saw:

$$\int_0^1x^{n+2}(1-x)^ndx = \left[\begin{align} t &=1-x \\-dt &=dx \\ x &=0,1\leftrightarrow t=1,0 \\x &=1-t\end{align} \right]=\int_0^1(1-t)^{n+2}t^ndt=\int_0^1(1-x)^{n+2}x^ndx$$

e.ad
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