What is the integral of (e^{-st})/s ?

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

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integral of (e^{-st})/s

\int \frac{e ^{- s t}}{s} d t

- e^{- s t} \frac{1}{s ^{2}} + C

Solution steps

\int \frac{e ^{- s t}}{s} d t

Take the constant out:

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

= \frac{1}{s} \cdot \int e^{- s t} d t

Apply u-substitution

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

= \frac{1}{s} \cdot \int - \frac{e ^{u}}{s} d u

Take the constant out:

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

= \frac{1}{s} (- \frac{1}{s} \cdot \int e^{u} d u)

Use the common integral:

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

= \frac{1}{s} (- \frac{1}{s} e^{u})

Substitute back

u = - s t

= \frac{1}{s} (- \frac{1}{s} e^{- s t})

Simplify

\frac{1}{s} (- \frac{1}{s} e^{- s t}) : - e^{- s t} \frac{1}{s ^{2}}

= - e^{- s t} \frac{1}{s ^{2}}

Add a constant to the solution

= - e^{- s t} \frac{1}{s ^{2}} + C

\int ln (6 x + 1) d x

\frac{d}{d x} (sin (e^{x} - 1))

4 x^{3} - 4 x^{2} + 3

n = 1 \sum \infty \frac{n ^{3}}{e ^{n}}

x \to 1 + lim (x^{\frac{8}{1 - x}})

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