If int {{e^{{x^2}}}left( {2 - frac{1}{{{x^2}} | vedclass

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(A) We are given the integral $\int e^{x^2} \left( 2 - \frac{1}{x^2} \right) dx = e^{x^2} f(x) + C$. Let $I = \int e^{x^2} \left( 2 - \frac{1}{x^2} \r

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(A) We are given the integral $\int e^{x^2} \left( 2 - \frac{1}{x^2} \right) dx = e^{x^2} f(x) + C$. Let $I = \int e^{x^2} \left( 2 - \frac{1}{x^2} \right) dx$. We can rewrite the integrand as: $e^{x^2} \left( 2 - \frac{1}{x^2} \right) = e^{x^2} \left( 2x \cdot \frac{1}{x} - \frac{1}{x^2} \right)$. Notice that if we let $g(x) = \frac{1}{x}$,then $g'(x) = -\frac{1}{x^2}$. Thus,the integral becomes $\int e^{x^2} (2x g(x) + g'(x)) dx$. This is not quite in the standard form $\int e^{h(x)} (h'(x)g(x) + g'(x)) dx$. Let's differentiate $e^{x^2} f(x)$ to see if it matches the integrand: $\frac{d}{dx} [e^{x^2} f(x)] = e^{x^2} (2x) f(x) + e^{x^2} f'(x) = e^{x^2} [2x f(x) + f'(x)]$. Comparing this to $e^{x^2} (2 - \frac{1}{x^2})$,we get $2x f(x) + f'(x) = 2 - \frac{1}{x^2}$. If we assume $f(x) = \frac{1}{x}$,then $f'(x) = -\frac{1}{x^2}$. Substituting these into the equation: $2x(\frac{1}{x}) + (-\frac{1}{x^2}) = 2 - \frac{1}{x^2}$,which matches. Thus,$f(x) = \frac{1}{x} + k$. Given $f(\frac{1}{2}) = 2$,we have $\frac{1}{1/2} + k = 2 \implies 2 + k = 2 \implies k = 0$. Therefore,$f(x) = \frac{1}{x}$. Finally,$f(1) = \frac{1}{1} = 1$.

If $\int {{e^{{x^2}}}\left( {2 - \frac{1}{{{x^2}}}} \right)dx = {e^{{x^2}}}f(x) + C} $ and $f\left( {\frac{1}{2}} \right) = 2$,then $f(1)$ is equal to (where $C$ is an arbitrary constant).

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