What is the integral of e^{at}e^{-st} ?

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

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integral of e^{at}e^{-st}

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

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

Solution steps

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

Simplify

e^{- s t} e^{a t} : e^{- s t + a t}

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

Apply u-substitution

= \int \frac{e ^{u}}{- s + a} d u

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - s t + a t

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

Add a constant to the solution

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

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