If \(\int \frac{x^2}{\sqrt{1-x}} \mathrm{~d} x=\mathrm{p} \sqrt{1-x}\left(3 x^2+4 x+8\right)+\mathrm{c}\) where \(\mathrm{c}\) is a constant of integration, then the value of \(\mathrm{p}\) is

\(\begin{aligned} & \text { Let } \mathrm{I}=\int \frac{x^2}{\sqrt{1-x}} \mathrm{~d} x \\ & \text { Put } 1-x=\mathrm{t} \\ & \Rightarrow \mathrm{d} x=-\mathrm{dt} \\ & \therefore \quad \mathrm{I}=\int \frac{(1-\mathrm{t})^2}{\sqrt{\mathrm{t}}} \cdot(-\mathrm{dt}) \\ & =-\int \frac{1-2 \mathrm{t}+\mathrm{t}^2}{\mathrm{t}^{\frac{1}{2}}} \mathrm{dt} \\ & =-\int\left(t^{-\frac{1}{2}}-2 t^{\frac{1}{2}}+t^{\frac{3}{2}}\right) d t \\ & =-\left(2 t^{\frac{1}{2}}-\frac{4}{3} t^{\frac{3}{2}}+\frac{2}{5} t^{\frac{5}{2}}\right)+c \\ & =t^{\frac{1}{2}}\left(\frac{-30+20 t-6 t^2}{15}\right)+c \\ & \end{aligned}\)
\(\begin{aligned} & =\sqrt{1-x}\left(\frac{-30+20(1-x)-6(1-x)^2}{15}\right)+\mathrm{c} \\ & =\sqrt{1-x}\left(\frac{-6 x^2-8 x-16}{15}\right)+\mathrm{c} \\ & =-\frac{2}{15} \sqrt{1-x}\left(3 x^2+4 x+8\right)+\mathrm{c} \\ \therefore \quad \mathrm{p} & =\frac{-2}{15}\end{aligned}\)