Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 28

Find the indefinite integral $\displaystyle \int \frac{x^2}{\sqrt{1 - x}} dx$

If we let $\displaystyle u = 1 - x$, then $\displaystyle du = - dx$, so $dx = - du$. Also, $x = 1 - u$, so $x^2 = (1 - u)^2$. Therefore,


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\begin{equation}
\begin{aligned}

\int \frac{x^2}{\sqrt{1 - x}} dx =& \int \frac{(1 - u)^2}{\sqrt{u}} - du
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\int \frac{x^2}{\sqrt{1 - x}} dx =& - \int \frac{(1 - u)^2}{\sqrt{u}} du
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\int \frac{x^2}{\sqrt{1 - x}} dx =& - \int \frac{1 - 2u + u^2}{\sqrt{u}} du
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\int \frac{x^2}{\sqrt{1 - x}} dx =& - \left( \int \frac{1}{\sqrt{u}} du - 2 \int \frac{u}{\sqrt{u}} du + \int \frac{u^2}{\sqrt{u}} du \right)
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\int \frac{x^2}{\sqrt{1 - x}} dx =& - \left( \int u^{\frac{-1}{2}} du - 2 \int u^{\frac{1}{2}} du + \int u^{\frac{3}{2}} du \right)
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\int \frac{x^2}{\sqrt{1 - x}} dx =& - \left[ \frac{u^{\frac{-1}{2}} + 1}{\displaystyle \frac{-1}{2} + 1} - 2 \left( \frac{u^{\frac{1}{2}} + 1}{\displaystyle \frac{1}{2}} + \right) + \frac{u^{\frac{3}{2}} + 1}{\displaystyle \frac{3}{2} + 1} \right] + C
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\int \frac{x^2}{\sqrt{1 - x}} dx =& - \left[ \frac{u^{\frac{1}{2}}}{\displaystyle \frac{1}{2}} - 2 \left( \frac{u^{\frac{3}{2}}}{\displaystyle \frac{3}{2}} \right) + \frac{u^{\frac{5}{2}}}{\displaystyle \frac{5}{2}} \right] + C
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\int \frac{x^2}{\sqrt{1 - x}} dx =& - \left( 2u^{\frac{1}{2}} - \frac{4 u^{\frac{3}{2}}}{3} + \frac{2 u^{\frac{5}{2}}}{5} \right) + C
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\int \frac{x^2}{\sqrt{1 - x}} dx =& -2 (1 - x)^{\frac{1}{2}} + \frac{4(1 - x)^{\frac{3}{2}}}{3} - \frac{2(1 - x)^{\frac{5}{2}}}{5} + C



\end{aligned}
\end{equation}
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