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

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

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integral of t(e^{-st})

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

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

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u} u}{s ^{2}} d u

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

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^{u} u d u

Apply Integration By Parts

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

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

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

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

Substitute back

u = - s t

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

Simplify

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

Add a constant to the solution

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

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\int \frac{3 x + 1}{( x - 2 ) ( x + 3 ) ^{2}} d x

\frac{\partial}{\partial x} (\frac{( y + z )}{x})

\frac{d}{d x} (k (sec (x) + tan (x)))

(4 + x) y^{^{'}} = 3 y

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