What is the integral of e^{-2/x} ?

The integral of e^{-2/x} is e^{-2/x}x+2\Ei(-2/x)+C

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integral of e^{-2/x}

\int e^{- \frac{2}{x}} d x

e^{- \frac{2}{x}} x + 2 Ei (- \frac{2}{x}) + C

Solution steps

\int e^{- \frac{2}{x}} d x

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integration By Parts

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

= - 2 (- \frac{e ^{- u}}{u} - \int \frac{e ^{- u}}{u} d u)

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

= - 2 (- \frac{e ^{- u}}{u} - Ei (- u))

Substitute back

u = \frac{2}{x}

= - 2 (- \frac{e ^{- \frac{2}{x}}}{\frac{2}{x}} - Ei (- \frac{2}{x}))

Simplify

- 2 (- \frac{e ^{- \frac{2}{x}}}{\frac{2}{x}} - Ei (- \frac{2}{x})) : e^{- \frac{2}{x}} x + 2 Ei (- \frac{2}{x})

= e^{- \frac{2}{x}} x + 2 Ei (- \frac{2}{x})

Add a constant to the solution

= e^{- \frac{2}{x}} x + 2 Ei (- \frac{2}{x}) + C

\int \frac{x ^{15}}{1 + x ^{16}} d x

\frac{d}{d x} ((\frac{x - 1}{x + 1})^{2})

\frac{d}{d x} (2 tan (2 x))

\frac{d}{d t} (cos^{3} (t))

\frac{d}{d x} ((6 x - 1) (3 x - 5)^{- 1})

What is the integral of e^{-2/x} ?
The integral of e^{-2/x} is e^{-2/x}x+2\Ei(-2/x)+C