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

The integral of 4e^{-2t}-te^{-2t} is 1/2 e^{-2t}t-7/4 e^{-2t}+C

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

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

\frac{1}{2} e^{- 2 t} t - \frac{7}{4} e^{- 2 t} + C

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

- 2 e^{- 2 t} - \frac{1}{4} (- 2 e^{- 2 t} t - e^{- 2 t}) : \frac{1}{2} e^{- 2 t} t - \frac{7}{4} e^{- 2 t}

= \frac{1}{2} e^{- 2 t} t - \frac{7}{4} e^{- 2 t}

Add a constant to the solution

= \frac{1}{2} e^{- 2 t} t - \frac{7}{4} e^{- 2 t} + C

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What is the integral of 4e^{-2t}-te^{-2t} ?
The integral of 4e^{-2t}-te^{-2t} is 1/2 e^{-2t}t-7/4 e^{-2t}+C