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

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

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integral of e^te^{-st}

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

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

Solution steps

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

Simplify

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

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

Apply u-substitution

: \int \frac{e ^{u}}{- s + 1} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - s t + t

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

Add a constant to the solution

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

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