What is the integral of 2e^{-2t}+4t+1 ?

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

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integral of 2e^{-2t}+4t+1

\int 2 e^{- 2 t} + 4 t + 1 d t

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

Solution steps

\int 2 e^{- 2 t} + 4 t + 1 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 2 e^{- 2 t} d t + \int 4 t d t + \int 1 d t

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

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

\int 1 d t = t

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

Add a constant to the solution

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

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