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}

integral

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

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

Solution steps

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

Simplify

= \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

e^{1 + x + l n (x)}

\frac{x ^{2} - 4}{x}

\frac{x ^{2}}{4}

(2 x + 1)^{3}

2 + \frac{1}{x ^{2}} - \frac{3}{x ^{4}}

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