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

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

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integral of e^{-st-t}

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - t - s t

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

Add a constant to the solution

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

\frac{d}{d x} (cos (x) x^{3})

\int (2 x + 9)^{5} d x

f (x) = \frac{4 x ^{2}}{x + 3}, (1, 1)

y = \frac{x}{x + 1}, (4, 0.4)

y^{^{''}} + 4 y = 5 csc^{2} (2 t)

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