What is the integral of t^3e^{-4t} ?

The integral of t^3e^{-4t} is -1/4 e^{-4t}t^3-3/128 e^{-4t}(8t^2+4t+1)+C

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integral of t^3e^{-4t}

\int t^{3} e^{- 4 t} d t

- \frac{1}{4} e^{- 4 t} t^{3} - \frac{3}{128} e^{- 4 t} (8 t^{2} + 4 t + 1) + C

Solution steps

\int t^{3} e^{- 4 t} d t

Apply Integration By Parts

: - \frac{1}{4} e^{- 4 t} t^{3} - \int - \frac{3}{4} e^{- 4 t} t^{2} d t

= - \frac{1}{4} e^{- 4 t} t^{3} - \int - \frac{3}{4} e^{- 4 t} t^{2} d t

\int - \frac{3}{4} e^{- 4 t} t^{2} d t = \frac{3}{128} e^{- 4 t} (8 t^{2} + 4 t + 1)

= - \frac{1}{4} e^{- 4 t} t^{3} - \frac{3}{128} e^{- 4 t} (8 t^{2} + 4 t + 1)

Add a constant to the solution

= - \frac{1}{4} e^{- 4 t} t^{3} - \frac{3}{128} e^{- 4 t} (8 t^{2} + 4 t + 1) + C

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What is the integral of t^3e^{-4t} ?
The integral of t^3e^{-4t} is -1/4 e^{-4t}t^3-3/128 e^{-4t}(8t^2+4t+1)+C