For $x < 1, \int \frac{x-x^2}{\sqrt{1-x}} d x=$

For $x < 1, \int \frac{x-x^2}{\sqrt{1-x}} d x=$
  1. $\frac{4}{3}(1-x)^{3 / 2}-\frac{2}{5}(1-x)^{5 / 2}-2 \sqrt{1-x}+c$
  2. $\frac{4}{3}(1-x)^{3 / 2}-\frac{2}{3}(1-x)^{5 / 2}-2 \sqrt{1-x}+c$
  3. $\frac{2}{3}(1-x)^{3 / 2}-2 \sqrt{1-x}+c$
  4. $-\frac{2}{15}(1-x)^{3 / 2}(2+3 x)+c$

Solution

$ \begin{aligned} & \text { For } x < 1, I=\int \frac{x-x^2}{\sqrt{1-x}} d x=\int \frac{x(1-x)}{\sqrt{1-x}} d x \\ & =\int x \sqrt{1-x} d x \end{aligned} $ Let $1-x=t^2$ $ \begin{array}{lc} \Rightarrow & d x=-2 t d t \\ \text { So, } & I=\int\left(1-t^2\right) t(-2 t) d t=2 \int\left(t^4-t^2\right) d t \\ \Rightarrow & I=2\left[\frac{t^5}{5}-\frac{t^3}{3}\right]+c \\ \Rightarrow & I=\frac{2}{15} t^3\left(3 t^2-5\right)+c \\ \Rightarrow & I=\frac{2}{15}(1-x)^{3 / 2}[3(1-x)-5]+c \\ \Rightarrow & I=\frac{-2}{15}(1-x)^{3 / 2}(3 x+2)+c . \end{array} $

Asked in: AP EAMCET 2018 (22 Apr Shift 2)

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