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

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

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integral of e^{1/y}

integral

e^{\frac{1}{y}}

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

Solution steps

\int e^{\frac{1}{y}} d y

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}{y}

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

Simplify

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

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

Add a constant to the solution

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

sin (- 2 x + 1)

t^{- 4}

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

\frac{1}{( x + 2 ) ^{2} \cdot ( \frac{1}{x} )}

(2 x)^{3}

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