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

The integral of e^{-st+t} is -e^{t-ts} 1/t+C

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integral of e^{-st+t}

\int e^{- s t + t}

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

Solution steps

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

e^{t - t s} = e^{t} e^{- t s}

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

Take the constant out:

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

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

Apply u-substitution

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

= e^{t} \cdot \int - \frac{e ^{u}}{t} d u

Take the constant out:

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

= e^{t} (- \frac{1}{t} \cdot \int e^{u} d u)

Use the common integral:

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

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

Substitute back

u = - t s

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

Simplify

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

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

Add a constant to the solution

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

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\int \frac{1}{( t - 1 ) ^{2}} d t

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What is the integral of e^{-st+t} ?
The integral of e^{-st+t} is -e^{t-ts} 1/t+C