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

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

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integral of (e^{1/y})/y

integral

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

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

Solution steps

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

Apply u-substitution

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

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

Use the nonelementary integral:

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

= - Ei (u)

Substitute back

u = \frac{1}{y}

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

Add a constant to the solution

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

\frac{1}{( sin ( x ) cos ^{2} ( x ) )}

x^{3} - x^{2}

(\frac{t - 1}{t})^{2}

\frac{1}{( x + 1 ) ^{3}}

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

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