What is the integral of (e^{-st})/(s^2) ?

The integral of (e^{-st})/(s^2) is -e^{-st} 1/(s^3)+C

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integral of (e^{-st})/(s^2)

\int \frac{e ^{- s t}}{s ^{2}} d t

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

Solution steps

\int \frac{e ^{- s t}}{s ^{2}} d t

Take the constant out:

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

= \frac{1}{s ^{2}} \cdot \int e^{- s t} d t

Apply u-substitution

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

= \frac{1}{s ^{2}} \cdot \int - \frac{e ^{u}}{s} d u

Take the constant out:

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

= \frac{1}{s ^{2}} (- \frac{1}{s} \cdot \int e^{u} d u)

Use the common integral:

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

= \frac{1}{s ^{2}} (- \frac{1}{s} e^{u})

Substitute back

u = - s t

= \frac{1}{s ^{2}} (- \frac{1}{s} e^{- s t})

Simplify

\frac{1}{s ^{2}} (- \frac{1}{s} e^{- s t}) : - e^{- s t} \frac{1}{s ^{3}}

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

Add a constant to the solution

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

y = \frac{100}{x ^{4}}

f (x) = 14400 + 600 x + x^{2}

\frac{\partial}{\partial x} (sin (2 x) cos (y))

\frac{d y}{d x} = e^{6 x} + 8 y

\frac{d y}{d x} = 5 x (x - 4)

What is the integral of (e^{-st})/(s^2) ?
The integral of (e^{-st})/(s^2) is -e^{-st} 1/(s^3)+C