Explanation

(c)
$I = \int_{0}^{1}{x(1 - x)^{n}dx}$
Put $1 - x = z \Rightarrow - dx = dz$
$\therefore I = \int_{1}^{0}{(1 - z)z^{n}( - dz)}$
$= \int_{0}^{1}{(z^{n} - z^{n + 1}})dz$
$= \left\lbrack \frac{z^{n = 1}}{n + 1} - \frac{z^{n + 2}}{n + 2} \right\rbrack_{0}^{1} = \frac{1}{n + 1} - \frac{1}{n + 2}$
Alternate Given integral can be rewritten as
$I = \int_{0}^{1}x^{2 - 1}(1 - x)^{(n + 1) - 1}dx$
It is the form of beta function
$\therefore\ \ I = B(2,n + 1) = \frac{\sqrt{2}\sqrt{n + 1}}{\sqrt{(n + 3}}$
$= \frac{1}{(n + 1)(n + 2)} = \frac{1}{n + 1} - \frac{1}{n + 2}$