What is the integral of e^{-st}(1) ?

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

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integral of e^{-st}(1)

\int e^{- s t} (1) d t

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

Solution steps

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

Simplify

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

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = - s t

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

Add a constant to the solution

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

f (t) = t^{2} - π^{2}

x \to 0 + lim (\frac{8}{5 x ^{\frac{1}{5}}})

\frac{d}{d x} (\frac{x - x}{x ^{\frac{1}{6}}})

\frac{d x}{d t} = e^{x} sin (t)

e^{2 s i n (3 v)}

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