What is the integral of e^{-st}t^3 ?

The integral of e^{-st}t^3 is 1/(s^4)(-s^3e^{-st}t^3-3(s^2e^{-st}t^2-2(-se^{-st}t-e^{-st})))+C

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integral of e^{-st}t^3

\int e^{- s t} t^{3} d t

\frac{1}{s ^{4}} (- s^{3} e^{- s t} t^{3} - 3 (s^{2} e^{- s t} t^{2} - 2 (- s e^{- s t} t - e^{- s t}))) + C

Solution steps

\int e^{- s t} t^{3} d t

Apply u-substitution

: \int \frac{e ^{u} u ^{3}}{s ^{4}} d u

= \int \frac{e ^{u} u ^{3}}{s ^{4}} d u

Take the constant out:

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

= \frac{1}{s ^{4}} \cdot \int e^{u} u^{3} d u

Apply Integration By Parts

: u^{3} e^{u} - \int 3 u^{2} e^{u} d u

= \frac{1}{s ^{4}} (u^{3} e^{u} - \int 3 u^{2} e^{u} d u)

\int 3 u^{2} e^{u} d u = 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

= \frac{1}{s ^{4}} (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = - s t

= \frac{1}{s ^{4}} ((- s t)^{3} e^{- s t} - 3 ((- s t)^{2} e^{- s t} - 2 (e^{- s t} (- s t) - e^{- s t})))

Simplify

\frac{1}{s ^{4}} ((- s t)^{3} e^{- s t} - 3 ((- s t)^{2} e^{- s t} - 2 (e^{- s t} (- s t) - e^{- s t}))) : \frac{1}{s ^{4}} (- s^{3} e^{- s t} t^{3} - 3 (s^{2} e^{- s t} t^{2} - 2 (- s e^{- s t} t - e^{- s t})))

= \frac{1}{s ^{4}} (- s^{3} e^{- s t} t^{3} - 3 (s^{2} e^{- s t} t^{2} - 2 (- s e^{- s t} t - e^{- s t})))

Add a constant to the solution

= \frac{1}{s ^{4}} (- s^{3} e^{- s t} t^{3} - 3 (s^{2} e^{- s t} t^{2} - 2 (- s e^{- s t} t - e^{- s t}))) + C

\frac{d}{d x} ((sin (x)) d x)

x \to 2 lim (\frac{x - 1}{x - 2})

x \to 0 + lim (cos^{\frac{1}{x}} (x))

\frac{\partial}{\partial x} (y^{5} - 3 x y)

x^{^{''}} + 4 x^{^{'}} + 4 x = 0

What is the integral of e^{-st}t^3 ?
The integral of e^{-st}t^3 is 1/(s^4)(-s^3e^{-st}t^3-3(s^2e^{-st}t^2-2(-se^{-st}t-e^{-st})))+C