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प्रश्न
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
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उत्तर
\[\int_{- a}^a \frac{x e^{x^2}}{1 + x^2} d x\]
\[\text{Let }f(x) = \frac{x e^{x^2}}{1 + x^2}\]
\[\text{Consider }f(-x) = - \frac{x e^{x^2}}{1 + x^2} = - f\left( x \right)\]
Thus f(x) is an odd function
Therefore,
\[ \int_{- a}^a \frac{x e^{x^2}}{1 + x^2} d x = 0\]
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