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

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(D) We know that $\int [f(x) + x f'(x)] dx = x f(x) + c$. Let $I = \int \left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} dx$. We can rewrite the integra

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$x e^{x+\frac{1}{x}}+c$,(where $c$ is a constant of integration)

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(D) We know that $\int [f(x) + x f'(x)] dx = x f(x) + c$. Let $I = \int \left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} dx$. We can rewrite the integrand as: $I = \int \left( x \left(1 - \frac{1}{x^2}\right) + 1 \right) e^{x+\frac{1}{x}} dx$. Let $f(x) = e^{x+\frac{1}{x}}$. Then $f'(x) = e^{x+\frac{1}{x}} \cdot \frac{d}{dx} \left(x + \frac{1}{x}\right) = e^{x+\frac{1}{x}} \left(1 - \frac{1}{x^2}\right)$. Substituting this into the integral,we get: $I = \int [x f'(x) + f(x)] dx$. Using the formula $\int [x f'(x) + f(x)] dx = x f(x) + c$,we obtain: $I = x e^{x+\frac{1}{x}} + c$.

$\int \left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} \,d x$ is equal to

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