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}

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

4 x - 2 x

lo g_{2} (e^{- x} cos (π x))

x \to \infty lim (r^{x})

x \to 0 + lim (1 - \frac{1}{x})

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