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

Take the constant out:

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

= - \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

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\int \frac{- 10 x ^{2}}{( x ^{3} + 2 ) ^{- 3}} d x

\frac{d}{d x} (\frac{250}{( 1 + 5 x ) ^{3}})

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