What is the integral of-12t^2e^{-3t} ?

The integral of-12t^2e^{-3t} is 4e^{-3t}t^2+8/3 e^{-3t}t+8/9 e^{-3t}+C

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integral of-12t^2e^{-3t}

\int - 12 t^{2} e^{- 3 t} d t

4 e^{- 3 t} t^{2} + \frac{8}{3} e^{- 3 t} t + \frac{8}{9} e^{- 3 t} + C

Solution steps

\int - 12 t^{2} e^{- 3 t} d t

Take the constant out:

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

= - 12 \cdot \int t^{2} e^{- 3 t} d t

Apply u-substitution

: \int - \frac{e ^{u} u ^{2}}{27} d u

= - 12 \cdot \int - \frac{e ^{u} u ^{2}}{27} d u

Take the constant out:

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

= - 12 (- \frac{1}{27} \cdot \int e^{u} u^{2} d u)

Apply Integration By Parts

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

= - 12 (- \frac{1}{27} (u^{2} e^{u} - \int 2 u e^{u} d u))

Take the constant out:

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

= - 12 (- \frac{1}{27} (u^{2} e^{u} - 2 \cdot \int u e^{u} d u))

Apply Integration By Parts

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

= - 12 (- \frac{1}{27} (u^{2} e^{u} - 2 (e^{u} u - \int e^{u} d u)))

Use the common integral:

\int e^{u} d u = e^{u}

= - 12 (- \frac{1}{27} (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = - 3 t

= - 12 (- \frac{1}{27} ((- 3 t)^{2} e^{- 3 t} - 2 (e^{- 3 t} (- 3 t) - e^{- 3 t})))

Simplify

- 12 (- \frac{1}{27} ((- 3 t)^{2} e^{- 3 t} - 2 (e^{- 3 t} (- 3 t) - e^{- 3 t}))) : 4 e^{- 3 t} t^{2} + \frac{8}{3} e^{- 3 t} t + \frac{8}{9} e^{- 3 t}

= 4 e^{- 3 t} t^{2} + \frac{8}{3} e^{- 3 t} t + \frac{8}{9} e^{- 3 t}

Add a constant to the solution

= 4 e^{- 3 t} t^{2} + \frac{8}{3} e^{- 3 t} t + \frac{8}{9} e^{- 3 t} + C

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What is the integral of-12t^2e^{-3t} ?
The integral of-12t^2e^{-3t} is 4e^{-3t}t^2+8/3 e^{-3t}t+8/9 e^{-3t}+C