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

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

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integral of e^{-st}(t)

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

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

Solution steps

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

Apply u-substitution

= \int \frac{e ^{u} u}{s ^{2}} d u

Take the constant out:

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

= \frac{1}{s ^{2}} \cdot \int e^{u} u d u

Apply Integration By Parts

= \frac{1}{s ^{2}} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{s ^{2}} (e^{u} u - e^{u})

Substitute back

u = - s t

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

Simplify

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

Add a constant to the solution

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

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