However, I am not sure how to arrive at this:

$$\begin{array}{rl}\phantom{=}& {{\int}_{0}^{1}{x}^{n+2}(1-x{)}^{n}\phantom{\rule{thinmathspace}{0ex}}dx}+2{\int}_{0}^{1}{x}^{n+1}(1-x{)}^{n+1}\phantom{\rule{thinmathspace}{0ex}}dx+{{\int}_{0}^{1}{x}^{n}(1-x{)}^{n+2}\phantom{\rule{thinmathspace}{0ex}}dx}\\ =& 2I(n+1)+{2{\int}_{0}^{1}{x}^{n+2}(1-x{)}^{n}\phantom{\rule{thinmathspace}{0ex}}dx}\end{array}$$

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