int left( {1 + x - frac{1}{x}} right){e^{x + | vedclass

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(D) Let $I = \int \left( {1 + x - \frac{1}{x}} \right){e^{x + \frac{1}{x}}} dx$. We can rewrite the integrand as: $I = \int \left( {e^{x + \frac{1}{x}

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$x{e^{x + \frac{1}{x}}} + C$

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(D) Let $I = \int \left( {1 + x - \frac{1}{x}} \right){e^{x + \frac{1}{x}}} dx$. We can rewrite the integrand as: $I = \int \left( {e^{x + \frac{1}{x}} + x \left( {1 - \frac{1}{{{x^2}}}} \right) {e^{x + \frac{1}{x}}}} \right) dx$. Let $f(x) = x e^{x + \frac{1}{x}}$. Then,by the product rule,$f'(x) = 1 \cdot e^{x + \frac{1}{x}} + x \cdot e^{x + \frac{1}{x}} \cdot \left( {1 - \frac{1}{{{x^2}}}} \right)$. $f'(x) = e^{x + \frac{1}{x}} + x \left( {1 - \frac{1}{{{x^2}}}} \right) e^{x + \frac{1}{x}}$. This matches the integrand. Therefore,$\int f'(x) dx = f(x) + C$. $I = x e^{x + \frac{1}{x}} + C$.

$\int \left( {1 + x - \frac{1}{x}} \right){e^{x + \frac{1}{x}}}\,dx = $

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