intleft(1+x-x^{-1}right) e^{x+x^{-1}} d x is | 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}}

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

Vedclass · Answer

(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 $\int e^{x + x^{-1}} dx$: Let $u = x$ and $dv = e^{x + x^{-1}} dx$. This is not straightforward. Instead, consider the derivative of $x e^{x + x^{-1}}$: $\frac{d}{dx} (x e^{x + x^{-1}}) = 1 \cdot e^{x + x^{-1}} + x \cdot e^{x + x^{-1}} \cdot (1 - x^{-2}) = e^{x + x^{-1}} + x e^{x + x^{-1}} - x^{-1} e^{x + x^{-1}} = (1 + x - x^{-1}) e^{x + x^{-1}}$. Therefore, $\int (1 + x - x^{-1}) e^{x + x^{-1}} dx = x e^{x + x^{-1}} + C$.

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

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