What is the integral of (e^{1/z})/(z^2) ?

The integral of (e^{1/z})/(z^2) is -e^{1/z}+C

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integral of (e^{1/z})/(z^2)

\int \frac{e ^{\frac{1}{z}}}{z ^{2}} d z

- e^{\frac{1}{z}} + C

Solution steps

\int \frac{e ^{\frac{1}{z}}}{z ^{2}} d z

Apply u-substitution

: \int - e^{u} d u

= \int - e^{u} d u

Take the constant out:

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

= - \int e^{u} d u

Use the common integral:

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

= - e^{u}

Substitute back

u = \frac{1}{z}

= - e^{\frac{1}{z}}

Add a constant to the solution

= - e^{\frac{1}{z}} + C

y = 3 - 4 x^{2}, (- 1, - 1)

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

\int 8 sin (x) cos (x) d x

\frac{d}{d x} (- 4 e^{2 x})

x \to 2 lim (\frac{1}{2 - x})

What is the integral of (e^{1/z})/(z^2) ?
The integral of (e^{1/z})/(z^2) is -e^{1/z}+C