What is the integral of e^{-t(a+s)} ?

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

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integral of e^{-t(a+s)}

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - (a + s) t

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

Add a constant to the solution

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

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