What is the integral of t^2e^{-18t} ?

The integral of t^2e^{-18t} is -1/2916 e^{-18t}(162t^2+18t+1)+C

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integral of t^2e^{-18t}

\int t^{2} e^{- 18 t} d t

- \frac{1}{2916} e^{- 18 t} (162 t^{2} + 18 t + 1) + C

Solution steps

\int t^{2} e^{- 18 t} d t

Apply u-substitution

: \int - \frac{e ^{u} u ^{2}}{5832} d u

= \int - \frac{e ^{u} u ^{2}}{5832} d u

Take the constant out:

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

= - \frac{1}{5832} \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}{5832} (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}{5832} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = - 18 t

= - \frac{1}{5832} ((- 18 t)^{2} e^{- 18 t} - 2 (e^{- 18 t} (- 18 t) - e^{- 18 t}))

Simplify

- \frac{1}{5832} ((- 18 t)^{2} e^{- 18 t} - 2 (e^{- 18 t} (- 18 t) - e^{- 18 t})) : - \frac{1}{2916} e^{- 18 t} (162 t^{2} + 18 t + 1)

= - \frac{1}{2916} e^{- 18 t} (162 t^{2} + 18 t + 1)

Add a constant to the solution

= - \frac{1}{2916} e^{- 18 t} (162 t^{2} + 18 t + 1) + C

g (x) = \frac{e ^{x}}{9 e ^{x} + x ^{2}}

\int \frac{x ^{2} - x + 12}{x ^{3} + 3 x} d x

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

\int 2 x^{- \frac{1}{3}} d x

y = x arcsin (x) + 1 - x^{2}

What is the integral of t^2e^{-18t} ?
The integral of t^2e^{-18t} is -1/2916 e^{-18t}(162t^2+18t+1)+C