Question

$\int \frac{x^2}{\sqrt{1-x}}dx$

Answer 1

$$\begin{gathered} \int \frac{x^2}{\sqrt{1-x}}dx \\ \mathrm{Let}1-x=t \\ \Rightarrow x=1-t \\ \Rightarrow 1=-\frac{dt}{dx} \\ \Rightarrow dx=-dt \\ \mathrm{Now},\int \frac{x^2}{\sqrt{1-x}}dx \\ =\int \frac{{\left(1-t\right)}^2}{\sqrt{t}}dt \\ =\int \left(\frac{1-t^2-2t}{\sqrt{t}}\right)dt \\ =\int \left(\frac{1}{\sqrt{t}}+\frac{t^2}{\sqrt{t}}-\frac{2t}{\sqrt{t}}\right)dt \\ =\int \left(t^{-\frac{1}{2}}+t^{\frac{3}{2}}-2t^{\frac{1}{2}}\right)dt \\ =\left[\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}+\frac{t^{{\displaystyle \frac{3}{2}}+1}}{{\displaystyle \frac{3}{2}}+1}-\frac{2t^{{\displaystyle \frac{1}{2}}+1}}{{\displaystyle \frac{1}{2}}+1}\right]+\mathrm{C} \\ =2t^{\frac{1}{2}}+\frac{2}{5}{t}^{\frac{5}{2}}-\frac{4}{3}{t}^{\frac{3}{2}}+\mathrm{C} \\ =2t^{\frac{1}{2}}\left[1+\frac{t^2}{5}-\frac{2}{3}t\right]+\mathrm{C} \\ =2t^{\frac{1}{2}}\left[\frac{15+3t^2-10t}{15}\right]+\mathrm{C} \\ =2\sqrt{1-x}\left[\frac{15+3{\left(1-x\right)}^2-10\left(1-x\right)}{15}\right]+\mathrm{C} \\ =\frac{2}{15}\sqrt{1-x}\left[15+3\left(1^2+x^2-2x\right)-10+10x\right]+\mathrm{C} \\ =\frac{2}{15}\sqrt{1-x}\left[15+3+3x^2-6x-10+10x\right]+\mathrm{C} \\ =\frac{2}{15}\sqrt{1-x}\left[3x^2+4x+8\right]+\mathrm{C} \end{gathered}$$