What is the integral of t^2e^{-6t} ?

The integral of t^2e^{-6t} is -1/108 e^{-6t}(18t^2+6t+1)+C

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integral of t^2e^{-6t}

\int t^{2} e^{- 6 t} d t

- \frac{1}{108} e^{- 6 t} (18 t^{2} + 6 t + 1) + C

Solution steps

\int t^{2} e^{- 6 t} d t

Apply u-substitution

: \int - \frac{e ^{u} u ^{2}}{216} d u

= \int - \frac{e ^{u} u ^{2}}{216} d u

Take the constant out:

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

= - \frac{1}{216} \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}{216} (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}{216} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = - 6 t

= - \frac{1}{216} ((- 6 t)^{2} e^{- 6 t} - 2 (e^{- 6 t} (- 6 t) - e^{- 6 t}))

Simplify

- \frac{1}{216} ((- 6 t)^{2} e^{- 6 t} - 2 (e^{- 6 t} (- 6 t) - e^{- 6 t})) : - \frac{1}{108} e^{- 6 t} (18 t^{2} + 6 t + 1)

= - \frac{1}{108} e^{- 6 t} (18 t^{2} + 6 t + 1)

Add a constant to the solution

= - \frac{1}{108} e^{- 6 t} (18 t^{2} + 6 t + 1) + C

n = 1 \sum \infty n!

\frac{d}{d x} (12 x^{3} + 6 x^{2} + 2 x)

y = cos (a^{5} + x^{5})

f (x) = 3 x^{2} + 2

\frac{7}{s ^{4}}

What is the integral of t^2e^{-6t} ?
The integral of t^2e^{-6t} is -1/108 e^{-6t}(18t^2+6t+1)+C