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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:
$x e^{x+x^{-1}}+C$
$\int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x$
$=\int e^{x+x^{-1}} d x+\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x$
$=e^{x+x^{-1}} \int d x-\int\left[\frac{d}{d x}\left(e^{x+x^{-1}}\right)\right] x d x$ $+\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x$
$=x e^{x+x^{-1}}-\int\left(1-\frac{1}{x^2}\right) x e^{x+x^{-1}} d x$ $+\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x$
$=x e^{x+x^{-1}}-\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x$
$+\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x+C$
$=x e^{x+x^{-1}}+C$
$=\int e^{x+x^{-1}} d x+\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x$
$=e^{x+x^{-1}} \int d x-\int\left[\frac{d}{d x}\left(e^{x+x^{-1}}\right)\right] x d x$ $+\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x$
$=x e^{x+x^{-1}}-\int\left(1-\frac{1}{x^2}\right) x e^{x+x^{-1}} d x$ $+\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x$
$=x e^{x+x^{-1}}-\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x$
$+\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x+C$
$=x e^{x+x^{-1}}+C$
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