What is the integral of e^{-st}2t ?

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

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integral of e^{-st}2t

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = - s t

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

Simplify

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

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

Add a constant to the solution

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

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