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

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Aicharya · Accepted Answer

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

If $\int \frac{x^2}{\sqrt{1-x}} \mathrm{~d} x=\mathrm{p} \sqrt{1-x}\left(3 x^2+4 x+8\right)+\mathrm{c}$…

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