Explanation

(a)
$\int_{}^{}\frac{x^{2}}{\sqrt{1 - x}}dx = p\sqrt{(1 - x)}(3x^{2} + 4x + 8)$
Put $1 - x = t^{2}$ in LHS, we get
$I = - \int_{}^{}\frac{2t}{t}\left( 1 - t^{2} \right)^{2}dt = - 2\int_{}^{}{(1 + t^{4} - 2t^{2})}dt$
$= - 2\left\lbrack t + \frac{t^{5}}{5} - \frac{2t^{3}}{3} \right\rbrack$
$= - 2\sqrt{1 - x}\left\lbrack 1 + \frac{(1 - x)^{2}}{5} - \frac{2}{3}(1 - x) \right\rbrack$
$= - 2\sqrt{1 - x}\left\lbrack \frac{15 + 3\left( 1 + x^{2} - 2x \right) - 10(1 - x)}{15} \right\rbrack$
$= \frac{- 2}{15}\sqrt{1 - x}(3x^{2} + 4x + 8)$
But $I = p\sqrt{1 - x}(3x^{2} + 4x + 8)$
$\therefore p = \frac{- 2}{15}$