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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int \left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} ~d x$. We can rewrite the integrand as: $I = \int \left[ x e^{x+\frac{1}{x}} \left(1-\frac{1}{x^2}\right) + e^{x+\frac{1}{x}} \right] d x$. This is in the form $\int [x f'(x) + f(x)] dx$,where $f(x) = e^{x+\frac{1}{x}}$. Since $\frac{d}{dx} (e^{x+\frac{1}{x}}) = e^{x+\frac{1}{x}} \cdot (1 - \frac{1}{x^2}) = f'(x)$,the integral evaluates to $x f(x) + c$. Therefore,$I = x e^{x+\frac{1}{x}} + c$.

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

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