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

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

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integral of 2te^{-st}

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

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

Solution steps

\int 2 t e^{- s 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

\frac{d}{d x} (3 x - sin^{2} (x))

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x \to 5 lim (x^{3})

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