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

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

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integral of 2e^{-t}+8

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

- 2 e^{- t} + 8 t + C

Solution steps

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

Apply u-substitution

= \int - 2 e^{u} - 8 d u

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^{u} d u - \int 8 d u

\int 2 e^{u} d u = 2 e^{u}

\int 8 d u = 8 u

= - 2 e^{u} - 8 u

Substitute back

u = - t

= - 2 e^{- t} - 8 (- t)

Simplify

= - 2 e^{- t} + 8 t

Add a constant to the solution

= - 2 e^{- t} + 8 t + C