What is the integral of e^{-(s+3)t} ?

The integral of e^{-(s+3)t} is 1/(-3-s)e^{-(3+s)t}+C

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integral of e^{-(s+3)t}

\int e^{- (s + 3) t} d t

\frac{1}{- 3 - s} e^{- (3 + s) t} + C

Solution steps

\int e^{- (s + 3) t} d t

Apply u-substitution

: \int \frac{e ^{u}}{- 3 - s} d u

= \int \frac{e ^{u}}{- 3 - s} d u

Take the constant out:

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

= \frac{1}{- 3 - s} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{- 3 - s} e^{u}

Substitute back

u = - (3 + s) t

= \frac{1}{- 3 - s} e^{- (3 + s) t}

Add a constant to the solution

= \frac{1}{- 3 - s} e^{- (3 + s) t} + C

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What is the integral of e^{-(s+3)t} ?
The integral of e^{-(s+3)t} is 1/(-3-s)e^{-(3+s)t}+C