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प्रश्न
\[\int\frac{x^2}{\sqrt{1 - x}} \text{ dx }\]
योग
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उत्तर
\[\text{We have}, \]
\[I = \int\frac{x^2}{\sqrt{1 - x}} dx\]
\[\text{ Let, 1 - x = t}^2 \]
\[\text{Differentiating both sides we get}\]
\[ - \text{ dx = 2t dt}\]
\[\text{Now, integration becomes}, \]
\[I = - \int\frac{\left( 1 - t^2 \right)^2}{t} 2tdt\]
\[ = - 2\int \left( 1 - t^2 \right)^2 dt\]
\[ = - 2\int\left( 1 - 2 t^2 + t^4 \right) dt\]
\[ = - 2\left[ t - \frac{2 t^3}{3} + \frac{t^5}{5} \right] + C\]
\[ = \frac{- 2}{15}t\left[ 3 t^4 - 10 t^2 + 15 \right] + C\]
\[ = \frac{- 2}{15}\sqrt{1 - x}\left[ 3 \left( 1 - x \right)^2 - 10\left( 1 - x \right) + 15 \right] + C\]
\[ = \frac{- 2}{15}\sqrt{1 - x}\left[ 3\left( 1 - 2x + x^2 \right) - 10\left( 1 - x \right) + 15 \right] + C\]
\[ = \frac{- 2}{15}\sqrt{1 - x}\left[ 3 x^2 - 6x + 3 - 10 + 10x + 15 \right] + C\]
\[ = \frac{- 2}{15}\sqrt{1 - x}\left[ 3 x^2 + 4x + 8 \right] + C\]
\[I = \int\frac{x^2}{\sqrt{1 - x}} dx\]
\[\text{ Let, 1 - x = t}^2 \]
\[\text{Differentiating both sides we get}\]
\[ - \text{ dx = 2t dt}\]
\[\text{Now, integration becomes}, \]
\[I = - \int\frac{\left( 1 - t^2 \right)^2}{t} 2tdt\]
\[ = - 2\int \left( 1 - t^2 \right)^2 dt\]
\[ = - 2\int\left( 1 - 2 t^2 + t^4 \right) dt\]
\[ = - 2\left[ t - \frac{2 t^3}{3} + \frac{t^5}{5} \right] + C\]
\[ = \frac{- 2}{15}t\left[ 3 t^4 - 10 t^2 + 15 \right] + C\]
\[ = \frac{- 2}{15}\sqrt{1 - x}\left[ 3 \left( 1 - x \right)^2 - 10\left( 1 - x \right) + 15 \right] + C\]
\[ = \frac{- 2}{15}\sqrt{1 - x}\left[ 3\left( 1 - 2x + x^2 \right) - 10\left( 1 - x \right) + 15 \right] + C\]
\[ = \frac{- 2}{15}\sqrt{1 - x}\left[ 3 x^2 - 6x + 3 - 10 + 10x + 15 \right] + C\]
\[ = \frac{- 2}{15}\sqrt{1 - x}\left[ 3 x^2 + 4x + 8 \right] + C\]
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