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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

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