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

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

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integral of 2t^2e^{-st}

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

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

Solution steps

\int 2 t^{2} 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^{2} d t

Apply Integration By Parts

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

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

\int - e^{- s t} \frac{2}{s} t d t = - \frac{2}{s ^{3}} (- s e^{- s t} t - e^{- s t})

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

Simplify

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

Add a constant to the solution

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

f (x) = 2 x^{3} - 4 x + 6, at x = 1

\int_{2}^{7} g (x) d x

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

\int e^{- a t} d t

\frac{d y}{d x} = \frac{x + y}{x + y + 1}

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