What is the integral of 2e^{-2t}+5e^{-8t} ?

The integral of 2e^{-2t}+5e^{-8t} is -e^{-2t}-5/8 e^{-8t}+C

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integral of 2e^{-2t}+5e^{-8t}

\int 2 e^{- 2 t} + 5 e^{- 8 t} d t

- e^{- 2 t} - \frac{5}{8} e^{- 8 t} + C

Solution steps

\int 2 e^{- 2 t} + 5 e^{- 8 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 2 e^{- 2 t} d t + \int 5 e^{- 8 t} d t

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

\int 5 e^{- 8 t} d t = - \frac{5}{8} e^{- 8 t}

= - e^{- 2 t} - \frac{5}{8} e^{- 8 t}

Add a constant to the solution

= - e^{- 2 t} - \frac{5}{8} e^{- 8 t} + C

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What is the integral of 2e^{-2t}+5e^{-8t} ?
The integral of 2e^{-2t}+5e^{-8t} is -e^{-2t}-5/8 e^{-8t}+C