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

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

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integral of e^{-1/x}

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

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

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u}}{u ^{2}} d u

= \int \frac{e ^{u}}{u ^{2}} d u

Apply Integration By Parts

: - \frac{e ^{u}}{u} - \int - \frac{e ^{u}}{u} d u

= - \frac{e ^{u}}{u} - \int - \frac{e ^{u}}{u} d u

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

= - \frac{e ^{u}}{u} - (- Ei (u))

Substitute back

u = - \frac{1}{x}

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

Simplify

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

\frac{\partial}{\partial x} (f (x, y) (x + y))

\int_{0}^{1} \frac{r ^{3}}{4 + r ^{2}} d r

x \to 2 lim (\frac{5 - x}{x - 2})

\frac{d}{d x} (\frac{sin ( x ) - x}{x ^{3}})

x \to \infty lim (4 e^{x})

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