int (1+x-x^{-1}) e^{x+x^{-1}} dx is equal to | vedclass

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

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$x e^{x+x^{-1}}+C$

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(D) Let $I = \int (1+x-x^{-1}) e^{x+x^{-1}} dx$. We can rewrite the integral as: $I = \int e^{x+x^{-1}} dx + \int (x-x^{-1}) e^{x+x^{-1}} dx$. Using integration by parts on the first term $\int e^{x+x^{-1}} dx$ by taking $u = x$ and $dv = e^{x+x^{-1}} dx$ is not direct,so let us consider the derivative of $x e^{x+x^{-1}}$: $\frac{d}{dx} (x e^{x+x^{-1}}) = 1 \cdot e^{x+x^{-1}} + x \cdot e^{x+x^{-1}} \cdot (1 - x^{-2}) = e^{x+x^{-1}} + x e^{x+x^{-1}} - x^{-1} e^{x+x^{-1}} = (1 + x - x^{-1}) e^{x+x^{-1}}$. Since the derivative of $x e^{x+x^{-1}}$ is exactly the integrand,the integral is $x e^{x+x^{-1}} + C$.

$\int (1+x-x^{-1}) e^{x+x^{-1}} dx$ is equal to :

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