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प्रश्न
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उत्तर १
\[\text{We have}, \]
\[I = \int\frac{e^x \left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[ = \int\frac{e^x \left( 1 + x^2 - 2x \right)}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[ = \int e^x \left[ \frac{\left( 1 + x^2 \right)}{\left( 1 + x^2 \right)^2} - \frac{2x}{\left( 1 + x^2 \right)^2} \right] \text{ dx }\]
\[ = \int e^x \left( \frac{1}{1 + x^2} - \frac{2x}{\left( 1 + x^2 \right)^2} \right) \text{ dx }\]
\[ = \frac{e^x}{1 + x^2} + C.......................[ ∵ \int e^x { f ( x ) + f' ( x ) } \text{ dx }= e^x f( x ) + C\]` \text{ Where} ,ƒ (x } = 1/{1+2} ⇒ ƒ ^' (x ) = - {2x}/ (1+ x^2)^2 ]`
उत्तर २
\[\text{We have}, \]
\[I = \int\frac{e^x \left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} dx\]
\[ = \int\frac{e^x \left( 1 + x^2 - 2x \right)}{\left( 1 + x^2 \right)^2} dx\]
\[ = \int e^x \left[ \frac{\left( 1 + x^2 \right)}{\left( 1 + x^2 \right)^2} - \frac{2x}{\left( 1 + x^2 \right)^2} \right] dx\]
\[ = \int e^x \left( \frac{1}{1 + x^2} - \frac{2x}{\left( 1 + x^2 \right)^2} \right) dx\]
\[ = \frac{e^x}{1 + x^2} + C \left[ \because \int e^x \left\{ f\left( x \right) + f'left( x \right) \right\} dx = e^x f\left( x \right) + C\]
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