What is the integral of 1/s e^{-st} ?

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

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integral of 1/s e^{-st}

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

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

Solution steps

\int \frac{1}{s} e^{- s t} 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

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