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$\int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x$ is equal to
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Verified Answer
The correct answer is:
$x e^{x+x^{-1}}+C$
$\int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x$
$=\int\left[x \cdot e^{x+x^{-1}}\left(1-\frac{1}{x^{2}}\right)+e^{x+x^{-1}}\right] d x$
$\left[\because \int x f^{\prime}(x)+f(x) d x=x f(x)+C\right]$
$\therefore \int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x=x e^{x+x^{-1}}+C$
$=\int\left[x \cdot e^{x+x^{-1}}\left(1-\frac{1}{x^{2}}\right)+e^{x+x^{-1}}\right] d x$
$\left[\because \int x f^{\prime}(x)+f(x) d x=x f(x)+C\right]$
$\therefore \int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x=x e^{x+x^{-1}}+C$
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