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

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

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

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

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

\frac{d}{d x} (\frac{2 x}{3 x ^{2} - 12})

(x^{3} + 4) (3 x^{2} - 4 x + 2)

3 t^{2} + 15 t + 19

\frac{d y}{d x} = \frac{x y - y}{y + 1}

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