If int(1+x-x^{-1}) e^{x+x^{-1}} dx = f(x)+c, | vedclass

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(B) Let $I = \int (1+x-x^{-1}) e^{x+x^{-1}} dx$. We can rewrite the integral as: $I = \int e^{x+x^{-1}} dx + \int x(1-x^{-2}) e^{x+x^{-1}} dx$. Observ

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$e^2+e^{-2}$

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(B) Let $I = \int (1+x-x^{-1}) e^{x+x^{-1}} dx$. We can rewrite the integral as: $I = \int e^{x+x^{-1}} dx + \int x(1-x^{-2}) e^{x+x^{-1}} dx$. Observe that the derivative of $e^{x+x^{-1}}$ is $(1-x^{-2}) e^{x+x^{-1}}$. Using the integration by parts formula $\int u dv = uv - \int v du$, let $u = x$ and $dv = (1-x^{-2}) e^{x+x^{-1}} dx$. Then $du = dx$ and $v = e^{x+x^{-1}}$. Thus, $\int x(1-x^{-2}) e^{x+x^{-1}} dx = x e^{x+x^{-1}} - \int e^{x+x^{-1}} dx$. Substituting this back into the expression for $I$: $I = \int e^{x+x^{-1}} dx + x e^{x+x^{-1}} - \int e^{x+x^{-1}} dx + c = x e^{x+x^{-1}} + c$. Therefore, $f(x) = x e^{x+x^{-1}}$. Now, calculate $f(1) - f(-1)$: $f(1) = 1 \cdot e^{1+1} = e^2$. $f(-1) = -1 \cdot e^{-1-1} = -e^{-2}$. $f(1) - f(-1) = e^2 - (-e^{-2}) = e^2 + e^{-2}$.

If $\int(1+x-x^{-1}) e^{x+x^{-1}} dx = f(x)+c$, then $f(1)-f(-1)=$

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