For x

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(D) Given $I = \int \frac{x-x^2}{\sqrt{1-x}} d x$. Since $x-x^2 = x(1-x)$,we have $I = \int \frac{x(1-x)}{\sqrt{1-x}} d x = \int x \sqrt{1-x} d x$. Le

Vedclass · Accepted Answer

$-\frac{2}{15}(1-x)^{3 / 2}(3x+2)+c$

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(D) Given $I = \int \frac{x-x^2}{\sqrt{1-x}} d x$. Since $x-x^2 = x(1-x)$,we have $I = \int \frac{x(1-x)}{\sqrt{1-x}} d x = \int x \sqrt{1-x} d x$. Let $1-x = t^2$,then $dx = -2t dt$. Substituting these into the integral: $I = \int (1-t^2) t (-2t) dt = 2 \int (t^4 - t^2) dt$. Integrating with respect to $t$: $I = 2 \left[ \frac{t^5}{5} - \frac{t^3}{3} \right] + c = \frac{2}{15} t^3 (3t^2 - 5) + c$. Substituting $t = \sqrt{1-x}$ back: $I = \frac{2}{15} (1-x)^{3/2} [3(1-x) - 5] + c = \frac{2}{15} (1-x)^{3/2} (-3x - 2) + c = -\frac{2}{15} (1-x)^{3/2} (3x+2) + c$.

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

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