What is the integral of t^3e^{-5t} ?

The integral of t^3e^{-5t} is -1/5 e^{-5t}t^3-3/625 (25e^{-5t}t^2-2(-5e^{-5t}t-e^{-5t}))+C

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integral of t^3e^{-5t}

\int t^{3} e^{- 5 t} d t

- \frac{1}{5} e^{- 5 t} t^{3} - \frac{3}{625} (25 e^{- 5 t} t^{2} - 2 (- 5 e^{- 5 t} t - e^{- 5 t})) + C

Solution steps

\int t^{3} e^{- 5 t} d t

Apply Integration By Parts

: - \frac{1}{5} e^{- 5 t} t^{3} - \int - \frac{3}{5} e^{- 5 t} t^{2} d t

= - \frac{1}{5} e^{- 5 t} t^{3} - \int - \frac{3}{5} e^{- 5 t} t^{2} d t

\int - \frac{3}{5} e^{- 5 t} t^{2} d t = \frac{3}{625} (25 e^{- 5 t} t^{2} - 2 (- 5 e^{- 5 t} t - e^{- 5 t}))

= - \frac{1}{5} e^{- 5 t} t^{3} - \frac{3}{625} (25 e^{- 5 t} t^{2} - 2 (- 5 e^{- 5 t} t - e^{- 5 t}))

Add a constant to the solution

= - \frac{1}{5} e^{- 5 t} t^{3} - \frac{3}{625} (25 e^{- 5 t} t^{2} - 2 (- 5 e^{- 5 t} t - e^{- 5 t})) + C

x \to 1 lim (e^{4 x^{3} - 4 x})

\frac{d}{d x} (((x + x^{9})^{- 6} + 3 x)^{3})

\frac{d}{d z} (x z)

4 y y^{^{'}} = x

What is the integral of t^3e^{-5t} ?
The integral of t^3e^{-5t} is -1/5 e^{-5t}t^3-3/625 (25e^{-5t}t^2-2(-5e^{-5t}t-e^{-5t}))+C