What is the integral of t^2e^{-5t} ?

The integral of t^2e^{-5t} is -1/125 (25e^{-5t}t^2-2(-5e^{-5t}t-e^{-5t}))+C

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integral of t^2e^{-5t}

\int t^{2} e^{- 5 t} d t

- \frac{1}{125} (25 e^{- 5 t} t^{2} - 2 (- 5 e^{- 5 t} t - e^{- 5 t})) + C

Solution steps

\int t^{2} e^{- 5 t} d t

Apply u-substitution

: \int - \frac{e ^{u} u ^{2}}{125} d u

= \int - \frac{e ^{u} u ^{2}}{125} d u

Take the constant out:

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

= - \frac{1}{125} \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}{125} (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}{125} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = - 5 t

= - \frac{1}{125} ((- 5 t)^{2} e^{- 5 t} - 2 (e^{- 5 t} (- 5 t) - e^{- 5 t}))

Simplify

- \frac{1}{125} ((- 5 t)^{2} e^{- 5 t} - 2 (e^{- 5 t} (- 5 t) - e^{- 5 t})) : - \frac{1}{125} (25 e^{- 5 t} t^{2} - 2 (- 5 e^{- 5 t} t - e^{- 5 t}))

= - \frac{1}{125} (25 e^{- 5 t} t^{2} - 2 (- 5 e^{- 5 t} t - e^{- 5 t}))

Add a constant to the solution

= - \frac{1}{125} (25 e^{- 5 t} t^{2} - 2 (- 5 e^{- 5 t} t - e^{- 5 t})) + C

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What is the integral of t^2e^{-5t} ?
The integral of t^2e^{-5t} is -1/125 (25e^{-5t}t^2-2(-5e^{-5t}t-e^{-5t}))+C