What is the integral of e^{-st}e^{-3t} ?

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

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integral of e^{-st}e^{-3t}

\int e^{- s t} e^{- 3 t} d t

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

Solution steps

\int e^{- s t} e^{- 3 t} d t

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{3 u ^{\frac{s + 3}{3}}}{s + 3}

= - \frac{1}{3} \cdot \frac{3 u ^{\frac{s + 3}{3}}}{s + 3}

Substitute back

u = e^{- 3 t}

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

Simplify

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

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

Add a constant to the solution

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

y = x, y = 14 x, x = 25

\frac{d y}{d x} = \frac{1}{x} (y^{2} + y)

f (x) = x^{2} - 7 x

\int y 10 + 4 y - 4 y^{2} d y

\frac{d y}{d x} = \frac{x + y}{x}

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