What is the integral of t^2e^{-4t} ?

The integral of t^2e^{-4t} is -1/32 e^{-4t}(8t^2+4t+1)+C

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integral of t^2e^{-4t}

\int t^{2} e^{- 4 t} d t

- \frac{1}{32} e^{- 4 t} (8 t^{2} + 4 t + 1) + C

Solution steps

\int t^{2} e^{- 4 t} d t

Apply u-substitution

: \int - \frac{e ^{u} u ^{2}}{64} d u

= \int - \frac{e ^{u} u ^{2}}{64} d u

Take the constant out:

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

= - \frac{1}{64} \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}{64} (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}{64} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = - 4 t

= - \frac{1}{64} ((- 4 t)^{2} e^{- 4 t} - 2 (e^{- 4 t} (- 4 t) - e^{- 4 t}))

Simplify

- \frac{1}{64} ((- 4 t)^{2} e^{- 4 t} - 2 (e^{- 4 t} (- 4 t) - e^{- 4 t})) : - \frac{1}{32} e^{- 4 t} (8 t^{2} + 4 t + 1)

= - \frac{1}{32} e^{- 4 t} (8 t^{2} + 4 t + 1)

Add a constant to the solution

= - \frac{1}{32} e^{- 4 t} (8 t^{2} + 4 t + 1) + C

\frac{d}{d x} (\frac{4 x}{7 x - 2})

\frac{1}{( 2 s + 1 ) s}

\int x^{2} ln (9 x) d x

\frac{d}{d x} (\frac{tan ^{2} ( x )}{tan ( x ^{2} )})

x \frac{d y}{d x} = e^{7 y + x}

What is the integral of t^2e^{-4t} ?
The integral of t^2e^{-4t} is -1/32 e^{-4t}(8t^2+4t+1)+C