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

(x + 1) e^{x + x^{- 1}} + C

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(x - 1) e^{x + x^{- 1}} + C

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

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

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Solution

The correct option is C $x e^{x + x^{- 1}} + C$
$\int (1 + x - x^{- 1}) e^{x + x^{- 1}} d x$

$= \int [x . e^{x + x^{- 1}} (1 - \frac{1}{x^{2}}) + e^{x + x^{- 1}}] d x$ $[∵ \int x f^{'} (x) + f (x) d x = x f (x) + C]$

$∴ \int (1 + x - x^{- 1}) e^{x + x^{- 1}} d x = x e^{x + x^{- 1}} + C$

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