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

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

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integral of (4e^{-2t}+4)

\int (4 e^{- 2 t} + 4) d t

- 2 e^{- 2 t} + 4 t + C

Solution steps

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

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

\int 4 d t = 4 t

= - 2 e^{- 2 t} + 4 t

Add a constant to the solution

= - 2 e^{- 2 t} + 4 t + C

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