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प्रश्न
\[\int \left( x + 1 \right)^2 e^x \text{ dx }\]
योग
उत्तर
\[\int \left( x + 1 \right)^2_I {e_{II}^x} \text{ dx }\]
\[ = \left( x + 1 \right)^2 \int e^x dx - \int\left\{ \frac{d}{dx} \left( x + 1 \right)^2 \int e^x dx \right\}dx\]
\[ = \left( x + 1 \right)^2 \cdot e^x - \int2 \left( x + 1 \right) \cdot e^x dx\]
\[ = \left( x + 1 \right)^2 e^x - 2\int \text{ x}_I {\text{ e}_{II}^x} \text{ dx} - 2\int e^x dx\]
\[ = \left( x + 1 \right)^2 e^x - 2 \left[ x \cdot e^x - \int1 \cdot e^x \text{ dx}\right] - 2 e^x \]
\[ = \left( x + 1 \right)^2 e^x - \text{ 2x e}^x + 2 e^x - 2 e^x + C\]
\[ = \left[ \left( x + 1 \right)^2 - 2x \right] e^x + C\]
\[ = \left( x^2 + 1 \right) e^x + C\]
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