What is the integral of e^{x^{-1}} ?

The integral of e^{x^{-1}} is e^{1/x}x-\Ei(1/x)+C

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integral of e^{x^{-1}}

\int e^{x^{- 1}} d x

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

Solution steps

\int e^{x^{- 1}} d x

Apply Integration By Parts

: e^{\frac{1}{x}} x - \int - \frac{e ^{\frac{1}{x}}}{x} d x

= e^{\frac{1}{x}} x - \int - \frac{e ^{\frac{1}{x}}}{x} d x

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

= e^{\frac{1}{x}} x - Ei (\frac{1}{x})

Add a constant to the solution

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

y = x^{3} - 7 x^{2} + 16 x + 7, at

\frac{d}{d x} (\frac{6 x}{7 - tan ( x )})

(- 8 cos (2 x))^{^{'}}

f (x) = 1 + x

\frac{d}{d x} (x (3 x + 2)^{4})

What is the integral of e^{x^{-1}} ?
The integral of e^{x^{-1}} is e^{1/x}x-\Ei(1/x)+C