What is the integral of t^2e^{-17t} ?

The integral of t^2e^{-17t} is -1/4913 (289e^{-17t}t^2-2(-17e^{-17t}t-e^{-17t}))+C

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integral of t^2e^{-17t}

\int t^{2} e^{- 17 t} d t

- \frac{1}{4913} (289 e^{- 17 t} t^{2} - 2 (- 17 e^{- 17 t} t - e^{- 17 t})) + C

Solution steps

\int t^{2} e^{- 17 t} d t

Apply u-substitution

: \int - \frac{e ^{u} u ^{2}}{4913} d u

= \int - \frac{e ^{u} u ^{2}}{4913} d u

Take the constant out:

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

= - \frac{1}{4913} \cdot \int e^{u} u^{2} d u

Apply Integration By Parts

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

= - \frac{1}{4913} (u^{2} e^{u} - \int 2 u e^{u} d u)

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

= - \frac{1}{4913} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = - 17 t

= - \frac{1}{4913} ((- 17 t)^{2} e^{- 17 t} - 2 (e^{- 17 t} (- 17 t) - e^{- 17 t}))

Simplify

- \frac{1}{4913} ((- 17 t)^{2} e^{- 17 t} - 2 (e^{- 17 t} (- 17 t) - e^{- 17 t})) : - \frac{1}{4913} (289 e^{- 17 t} t^{2} - 2 (- 17 e^{- 17 t} t - e^{- 17 t}))

= - \frac{1}{4913} (289 e^{- 17 t} t^{2} - 2 (- 17 e^{- 17 t} t - e^{- 17 t}))

Add a constant to the solution

= - \frac{1}{4913} (289 e^{- 17 t} t^{2} - 2 (- 17 e^{- 17 t} t - e^{- 17 t})) + C

x \to 3 lim (9 x + 92)

\frac{d}{d x} ((f (x))^{n})

\int e^{- 18 x} d x

\frac{\partial}{\partial t} (s - t)

f (x) = - 8 x^{3} + 24 x + 9

What is the integral of t^2e^{-17t} ?
The integral of t^2e^{-17t} is -1/4913 (289e^{-17t}t^2-2(-17e^{-17t}t-e^{-17t}))+C