The integral ∫(1+x−x1)ex+x1dx is equal to

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The integral {({1}+{x}-{{1}}{{x}})}{e}^{{{x}+{{1}}{{x}}}}{.{d}

Question

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

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

x e^{x + \frac{1}{x}} + C

(x + 1) e^{x + \frac{1}{x}} + C

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

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Liam Discussed

The integral

\int (1 + x - \frac{1}{x}) e^{x + \frac{1}{x}} d x

is equal to

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Question Text	The integral $\int (1 + x - \frac{1}{x}) e^{x + \frac{1}{x}} d x$ is equal to
Answer Type	Video solution: 1
Upvotes	150